chapter 11 systems of equations and inequalities...the solution of the system is 11, 36 xy== or −...

1026

Copyright © 2013 Pearson Education, Inc.

Chapter 11 Systems of Equations and Inequalities

Section 11.1 1. 3 4 8

4 4

1

x x

x

x

+ = −==

The solution set is { }1 .

2. a. 3 4 12x y+ =

x-intercept: ( )3 4 0 12

3 12

4

x

x

x

+ ===

y-intercept: ( )3 0 4 12

4 12

3

y

y

y

+ ===

b. 3 4 12

4 3 12

33

4

x y

y x

y x

+ == − +

= − +

A parallel line would have slope 34

− .

3. inconsistent

4. consistent; independent

5. (3, 2)−

6. consistent; dependent

7. 2 5

5 2 8

x y

x y

− =+ =

Substituting the values of the variables: 2(2) ( 1) 4 1 5

5(2) 2( 1) 10 2 8

− − = + =+ − = − =

Each equation is satisfied, so 2, 1x y= = − , or

(2, 1)− , is a solution of the system of equations.

8. 3 2 2

7 30

x y

x y

+ =− = −

Substituting the values of the variables: 3( 2) 2(4) 6 8 2

( 2) 7(4) 2 28 30

− + = − + =− − = − − = −

Each equation is satisfied, so 2, 4x y= − = , or

( 2, 4)− , is a solution of the system of equations.

9. 3 4 4

1 13

2 2

x y

x y

− =

− = −

Substituting the values of the variables: 1

3(2) 4 6 2 42

1 1 3 1(2) 3 1

2 2 2 2

− = − =

− = − = −

Each equation is satisfied, so 12, 2x y= = , or

( )12, 2 , is a solution of the system of equations.

10. 12 02

193 4 2

x y

x y

+ =

− = −

Substituting the values of the variables, we obtain:

( )

( )

1 12 2 1 1 0

2 2

1 3 193 4 2 8

2 2 2

− + = − + =

− − = − − = −

Each equation is satisfied, so 1 , 22x y= − = , or

( )1 , 22− , is a solution of the system of equations.

11. 3

13

2

x y

x y

− =

+ =

Substituting the values of the variables, we obtain:

Section 11.1: Systems of Linear Equations: Substitution and Elimination

1027


4 1 3

1(4) 1 2 1 3

2

− =

+ = + =

Each equation is satisfied, so 4, 1x y= = , or

(4, 1) , is a solution of the system of equations.

12. 3

3 1

x y

x y

− =− + =

Substituting the values of the variables:

( ) ( )( ) ( )2 5 2 5 3

3 2 5 6 5 1

− − − = − + =− − + − = − =

Each equation is satisfied, so 2, 5x y= − = − , or

( 2, 5)− − , is a solution of the system of equations.

13. 3 3 2 4

0

2 3 8

x y z

x y z

y z

+ + =− − =

− = −

Substituting the values of the variables: 3(1) 3( 1) 2(2) 3 3 4 4

1 ( 1) 2 1 1 2 0

2( 1) 3(2) 2 6 8

+ − + = − + =− − − = + − =

− − = − − = −

Each equation is satisfied, so 1, 1, 2x y z= = − = ,

or (1, 1, 2)− , is a solution of the system of equations.

14. 4 7

8 5 0

5 6

x z

x y z

x y z

− =+ − =

− − + =

Substituting the values of the variables: ( )

( ) ( )( ) ( )

4 2 1 8 1 7

8 2 5 3 1 16 15 1 0

2 3 5 1 2 3 5 6

− = − =+ − − = − − =

− − − + = − + + =

Each equation is satisfied, so 2x = , 3y = − ,

1z = , or (2, 3,1)− , is a solution of the system of equations.

15. 3 3 2 4

3 10

5 2 3 8

x y z

x y z

x y z

+ + =− + =

− − =

Substituting the values of the variables: ( ) ( ) ( )

( )( ) ( ) ( )

3 2 3 2 2 2 6 6 4 4

2 3 2 2 2 6 2 10

5 2 2 2 3 2 10 4 6 8

+ − + = − + =− − + = + + =

− − − = + − =


2z = , or (2, 2, 2)− is a solution of the system of equations.

16. 4 5 6

5 17

6 5 24

x z

y z

x y z

− =− = −

− − + =

Substituting the values of the variables: ( ) ( )( ) ( )( ) ( ) ( )

4 4 5 2 16 10 6

5 3 2 15 2 17

4 6 3 5 2 4 18 10 24

− = − =− − = − − = −

− − − + = − + + =


2z = , or (4, 3, 2)− , is a solution of the system of equations.

17. 8

4

x y

x y

+ =− =

Solve the first equation for y, substitute into the second equation and solve:

8

4

y x

x y

= −− =

(8 ) 4

8 4

2 12

6

x x

x x

x

x

− − =− + =

==

Since 6, 8 6 2x y= = − = . The solution of the system is 6, 2x y= = or using ordered pairs

(6, 2) .

18. 2 7

3

x y

x y

+ = −+ = −

Solve the first equation for x, substitute into the second equation and solve:

7 2

3

x y

x y

= − −+ = −

( 7 2 ) 3

7 3

4

y y

y

y

− − + = −− − = −

− =

Since 4, 7 2( 4) 1y x= − = − − − = . The solution of the system is 1, 4x y= = − or using an

ordered pair, (1, 4)− .

Chapter 11: Systems of Equations and Inequalities

1028 Copyright © 2013 Pearson Education, Inc.

19. 5 21

2 3 12

x y

x y

− =+ = −

Multiply each side of the first equation by 3 and add the equations to eliminate y:

15 3 63

2 3 12

x y

x y

− =+ = −

17 51

3

x

x

==

Substitute and solve for y: 5(3) 21

15 21

6

6

y

y

y

y

− =− =− =

= −

The solution of the system is 3, 6x y= = − or

using ordered an pair ( )3, 6− .

20. 3 5

2 3 8

x y

x y

+ =− = −

Add the equations: 3 5

2 3 8

x y

x y

+ =− = −

3 3

1

x

x

= −= −

Substitute and solve for y: 1 3 5

3 6

2

y

y

y

− + ===

The solution of the system is 1, 2x y= − = or

using ordered pairs ( 1, 2)− .

21. 3 24

2 0

x

x y

=+ =

Solve the first equation for x and substitute into the second equation:

8

2 0

x

x y

=+ =

8 2 0

2 8

4

y

y

y

+ == −= −


using ordered pairs (8, 4)−

22. 4 5 3

2 8

x y

y

+ = −− = −

Solve the second equation for y and substitute into the first equation:

4 5 3

4

x y

y

+ = −=

4 5(4) 3

4 20 3

4 23

234

x

x

x

x

+ = −+ = −

= −

= −

The solution of the system is 23

, 44

x y= − = or

using ordered pairs 23

, 44

− .

23. 3 6 2

5 4 1

x y

x y

− =+ =

Multiply each side of the first equation by 2 and each side of the second equation by 3, then add to eliminate y:

6 12 4

15 12 3

x y

x y

− =+ =

21 7

13

x

x

=

=

Substitute and solve for y: ( )3 1/ 3 6 2

1 6 2

6 1

16

y

y

y

y

− =− =− =

= −

The solution of the system is 1 1

,3 6

x y= = − or

using ordered pairs 1 1

,3 6

− .


1029


24. 2

2 43

3 5 10

x y

x y

+ =

− = −

Multiply each side of the first equation by 5 and each side of the second equation by 4, then add to eliminate y:

1010 20

312 20 40

x y

x y

+ =

− = −

11022

353

x

x

=

= −

Substitute and solve for y: ( )3 5 / 3 5 10

5 5 10

5 5

1

y

y

y

y

− − = −− − = −

− = −=


, 13

x y= − = or


, 13

− .

25. 2 1

4 2 3

x y

x y

+ =+ =


1 2

4 2 3

y x

x y

= −+ =

4 2(1 2 ) 3

4 2 4 3

0 1

x x

x x

+ − =+ − =

=

This equation is false, so the system is inconsistent.

26. 5

3 3 2

x y

x y

− =− + =


5

3 3 2

x y

x y

= +− + =

3( 5) 3 2

3 15 3 2

0 17

y y

y y

− + + =− − + =

=


27. 2 04 2 12

x yx y

− =+ =


24 2 12y xx y=+ =

4 2(2 ) 124 4 12

8 1232

x xx x

x

x

+ =+ =

=

=

Since 3 3

, 2 32 2

x y= = =


, 32

x y= = or


,3 .2

28. 3 3 1

84

3

x y

x y

+ = −

+ =

Solve the second equation for y, substitute into the first equation and solve:

3 3 18

43

x y

y x

+ = −

= −

8

3 3 4 13

3 8 12 19 9

1

x x

x xxx

+ − = −

+ − = −− = −

=

Since 8 8 4

1, 4(1) 4 .3 3 3

x y= = − = − = −


1,3

x y= = − or


1,3

− .

29. 2 4

2 4 8x yx y

+ =+ =


4 22 4 8x y

x y= −+ =



2(4 2 ) 4 88 4 4 8

0 0

y yy y

− + =− + =

=

These equations are dependent. The solution of the system is either 4 2x y= − , where y is any real

number or 4

2x

y−= , where x is any real number.

Using ordered pairs, we write the solution as

{ }( , ) 4 2 , is any real numberx y x y y= − or as

4( , ) , is any real number

2x

x y y x−= .

30. 3 7

9 3 21

x y

x y

− =− =


3 7

9 3 21

y x

x y

= −− =

9 3(3 7) 21

9 9 21 21

0 0

x x

x x

− − =− + =

=

These equations are dependent. The solution of the system is either 3 7y x= − , where x is any real

number is 7

3y

x+= , where y is any real number.

Using ordered pairs, we write the solution as

{ }( , ) 3 7, is any real numberx y y x x= − or as

7( , ) , is any real number

3y

x y x y+= .

31. 2 3 1

10 11

x y

x y

− = −+ =

Multiply each side of the first equation by –5, and add the equations to eliminate x:

10 15 5

10 11

x y

x y

− + =+ =

16 16

1

y

y

==

Substitute and solve for x: 2 3(1) 1

2 3 1

2 2

1

x

x

x

x

− = −− = −

==

The solution of the system is 1, 1x y= = or using ordered pairs (1, 1).

32. 3 2 0

5 10 4

x y

x y

− =+ =

Multiply each side of the first equation by 5, and add the equations to eliminate y:

15 10 0

5 10 4

x y

x y

− =+ =

20 4

15

x

x

=

=

Substitute and solve for y: ( )5 1/ 5 10 4

1 10 4

10 3

310

y

y

y

y

+ =+ =

=

=


,5 10

x y= = or


,5 10

.

33. 2 3 6

12

x y

x y

+ =

− =

Solve the second equation for x, substitute into the first equation and solve:

2 3 6

12

x y

x y

+ =

= +

12 3 6

22 1 3 6

5 5

1

y y

y y

y

y

+ + =

+ + ===

Since 1 3

1, 12 2

y x= = + = . The solution of the

system is 3

, 12

x y= = or using ordered pairs

3, 1

2.


1031


34. 1

22

2 8

x y

x y

+ = −

− =

Solve the second equation for x, substitute into the first equation and solve:

12

22 8

x y

x y

+ = −

= +

1(2 8) 2

24 2

2 6

3

y y

y y

y

y

+ + = −

+ + = −= −= −

Since 3, 2( 3) 8 6 8 2y x= − = − + = − + = . The solution of the system is 2, 3x y= = − or using

ordered pairs (2, 3)− .

35.

1 13

2 31 2

14 3

x y

x y

+ =

− = −

Multiply each side of the first equation by –6 and each side of the second equation by 12, then add to eliminate x:

3 2 18

3 8 12

x y

x y

− − = −− = −

10 30

3

y

y

− = −=

Substitute and solve for x: 1 1

(3) 32 3

11 3

21

22

4

x

x

x

x

+ =

+ =

=

=


36.

1 35

3 23 1

114 3

x y

x y

− = −

+ =

Multiply each side of the first equation by –54 and each side of the second equation by 24, then add to eliminate x:

18 81 270

18 8 264

x y

x y

− + =+ =

89 534

6

y

y

==

Substitute and solve for x: 3 1

(6) 114 3

32 11

43

94

12

x

x

x

x

+ =

+ =

=

=


37. 3 5 3

15 5 21

x y

x y

− =+ =

Add the equations to eliminate y: 3 5 3

15 5 21

x y

x y

− =+ =

18 24

43

x

x

=

=

Substitute and solve for y: ( )3 4 / 3 5 3

4 5 3

5 1

15

y

y

y

y

− =− =− = −

=


,3 5

x y= = or


,3 5

.



38. 2 1

1 32 2

x y

x y

− = −

+ =

Multiply each side of the second equation by 2, and add the equations to eliminate y:

2 1

2 3

x y

x y

− = −+ =

4 2

12

x

x

=

=

Substitute and solve for y: 1

2 12

1 1

2

2

y

y

y

y

− = −

− = −− = −

=


, 22

x y= = or


, 22

.

39.

1 18

3 50

x y

x y

+ =

− =

Rewrite letting 1 1

,u vx y

= = :

8

3 5 0

u v

u v

+ =− =

Solve the first equation for u, substitute into the second equation and solve:

8

3 5 0

u v

u v

= −− =

3(8 ) 5 0

24 3 5 0

8 24

3

v v

v v

v

v

− − =− − =

− = −=

Since 3, 8 3 5v u= = − = . Thus, 1 1

5x

u= = ,

1 13

yv

= = . The solution of the system is

1 1,

5 3x y= = or using ordered pairs

1 1,

5 3.

40.

4 30

6 32

2

x y

x y

− =

+ =

Rewrite letting 1 1

,u vx y

= = :

4 3 0

36 2

2

u v

u v

− =

+ =

Multiply each side of the second equation by 2, and add the equations to eliminate v:

4 3 0

12 3 4

u v

u v

− =+ =

16 4

4 116 4

u

u

=

= =

Substitute and solve for v: 1

4 3 04

1 3 0

3 1

13

v

v

v

v

− =

− =− = −

=

Thus, 1 1

4, 3x yu v

= = = = . The solution of the

system is 4, 3x y= = or using ordered pairs (4, 3).

41. 6

2 3 16

2 4

x y

x z

y z

− =− =+ =

Multiply each side of the first equation by –2 and add to the second equation to eliminate x:

2 2 12

2 3 16

2 3 4

x y

x z

y z

− + = −− =− =

Multiply each side of the result by –1 and add to the original third equation to eliminate y:

2 3 4

2 4

4 0

0

y z

y z

z

z

− + = −+ =

==

Substituting and solving for the other variables:


1033


2 0 4

2 4

2

y

y

y

+ ===

2 3(0) 16

2 16

8

x

x

x

− ===

The solution is 8, 2, 0x y z= = = or using ordered triples (8, 2, 0).

42. 2 4

2 4 0

3 2 11

x y

y z

x z

+ = −− + =

− = −

Multiply each side of the first equation by 2 and add to the second equation to eliminate y: 4 2 8

2 4 0

4 4 8

x y

y z

x z

+ = −− + =

+ = −

Multiply each side of the result by 12

and add to

the original third equation to eliminate z: 2 2 4

3 2 11

5 15

3

x z

x z

x

x

+ = −− = −

= −= −

Substituting and solving for the other variables: 2( 3) 4

6 4

2

y

y

y

− + = −− + = −

=

3( 3) 2 11

9 2 11

2 2

1

z

z

z

z

− − = −− − = −

− = −=

The solution is 3, 2, 1x y z= − = = or using

ordered triples ( 3, 2,1)− .

43. 2 3 7

2 43 2 2 10

x y zx y zx y z

− + =+ + =

− + − = −

Multiply each side of the first equation by –2 and add to the second equation to eliminate x; and multiply each side of the first equation by 3 and add to the third equation to eliminate x:

2 4 6 14

2 4

5 5 10

x y z

x y z

y z

− + − = −+ + =

− = −

3 6 9 21

3 2 2 10

4 7 11

x y z

x y z

y z

− + =− + − = −

− + =

Multiply each side of the first result by 45

and

add to the second result to eliminate y:

4 4 8

4 7 11

y z

y z

− = −− + =

3 3

1

z

z

==


1 2

1

y

y

− = −= −

2( 1) 3(1) 7

2 3 7

2

x

x

x

− − + =+ + =

=

The solution is 2, 1, 1x y z= = − = or using

ordered triples (2, 1,1)− .

44. 2 3 0

2 2 7

3 4 3 7

x y z

x y z

x y z

+ − =− + + = −

− − =

Multiply each side of the first equation by –2 and add to the second equation to eliminate y; and multiply each side of the first equation by 4 and add to the third equation to eliminate y:

4 2 6 0

2 2 7

6 7 7

x y z

x y z

x z

− − + =− + + = −− + = −

8 4 12 0

3 4 3 7

11 15 7

x y z

x y z

x z

+ − =− − =

− =

Multiply each side of the first result by 11 and multiply each side of the second result by 6 to eliminate x:

66 77 77

66 90 42

x z

x z

− + = −− =

13 35

3513

z

z

− = −

=

Substituting and solving for the other variables: 35

6 7 713245

6 713

3366

135613

x

x

x

x

− + = −

− + = −

− = −

=



56 352 3 0

13 13112 105

013 13

713

y

y

y

+ − =

+ − =

= −

The solution is 56 7 35

, ,13 13 13

x y z= = − = or

using ordered triples 56 7 35

, ,13 13 13

− .

45. 1

2 3 2

3 2 0

x y z

x y z

x y

− − =+ + =+ =

Add the first and second equations to eliminate z: 1

2 3 2

3 2 3

x y z

x y z

x y

− − =+ + =+ =

Multiply each side of the result by –1 and add to the original third equation to eliminate y:

3 2 3

3 2 0

0 3

x y

x y

− − = −+ =

= −


46. 2 3 0

2 5

3 4 1

x y z

x y z

x y z

− − =− + + =

− − =

Add the first and second equations to eliminate z; then add the second and third equations to eliminate z: 2 3 0

2 5

5

x y z

x y z

x y

− − =− + + =

− =

2 5

3 4 1

2 2 6

x y z

x y z

x y

− + + =− − =

− =

Multiply each side of the first result by –2 and add to the second result to eliminate y:

2 2 10

2 2 6

x y

x y

− + = −− =

0 2= − This equation is false, so the system is inconsistent.

47. 1

2 3 4

3 2 7 0

x y z

x y z

x y z

− − =− + − = −

− − =

Add the first and second equations to eliminate x; multiply the first equation by –3 and add to the third equation to eliminate x:

1

2 3 4

4 3

x y z

x y z

y z

− − =− + − = −

− = −

3 3 3 3

3 2 7 0

4 3

x y z

x y z

y z

− + + = −− − =

− = −


4 3

4 3

0 0

y z

y z

− + =− = −

=

The system is dependent. If z is any real number, then 4 3y z= − . Solving for x in terms of z in the first equation:

(4 3) 1

4 3 1

5 3 1

5 2

x z z

x z z

x z

x z

− − − =− + − =

− + == −

The solution is {( , , )x y z 5 2,x z= − 4 3y z= − ,

z is any real number}.

48. 2 3 0

3 2 2 2

5 3 2

x y z

x y z

x y z

− − =+ + =+ + =

Multiply the first equation by 2 and add to the second equation to eliminate z; multiply the first equation by 3 and add to the third equation to eliminate z: 4 6 2 0

3 2 2 2

7 4 2

x y z

x y z

x y

− − =+ + =− =

6 9 3 0

5 3 2

7 4 2

x y z

x y z

x y

− − =+ + =− =



1035


7 4 2

7 4 2

0 0

x y

x y

− + = −− =

=

The system is dependent. If y is any real

number, then 4 27 7

x y= + .

Solving for z in terms of x in the first equation: 2 3

4 22 3

78 4 21

713 4

7

z x y

yy

y y

y

= −+= −

+ −=

− +=

The solution is 4 2

( , , )7 7

x y z x y= + ,

13 4, is any real number

7 7z y y= − + .

49. 2 2 3 6

4 3 2 0

2 3 7 1

x y z

x y z

x y z

− + =− + =

− + − =

Multiply the first equation by –2 and add to the second equation to eliminate x; add the first and third equations to eliminate x:

4 4 6 12

4 3 2 0

4 12

x y z

x y z

y z

− + − = −− + =

− = −

2 2 3 6

2 3 7 1

4 7

x y z

x y z

y z

− + =− + − =

− =


4 12

4 7

0 19

y z

y z

− + =− =

=

This result is false, so the system is inconsistent.

50. 3 2 2 6

7 3 2 1

2 3 4 0

x y z

x y z

x y z

− + =− + = −− + =

Multiply the first equation by –1 and add to the second equation to eliminate z; multiply the first equation by –2 and add to the third equation to eliminate z:

3 2 2 6

7 3 2 1

4 7

x y z

x y z

x y

− + − = −− + = −− = −

6 4 4 12

2 3 4 0

4 12

x y z

x y z

x y

− + − = −− + =

− + = −

Add the first result to the second result to eliminate y:

4 7

4 12

0 19

x y

x y

− = −− + = −

= −

This result is false, so the system is inconsistent.

51. 6

3 2 5

3 2 14

x y z

x y z

x y z

+ − =− + = −+ − =

Add the first and second equations to eliminate z; multiply the second equation by 2 and add to the third equation to eliminate z:

6

3 2 5

4 1

x y z

x y z

x y

+ − =− + = −− =

6 4 2 10

3 2 14

7 4

x y z

x y z

x y

− + = −+ − =− =


4 1

7 4

x y

x y

− + = −− =

3 3

1

x

x

==

Substituting and solving for the other variables: 4(1) 1

3

3

y

y

y

− =− = −

=

3(1) 2(3) 5

3 6 5

2

z

z

z

− + = −− + = −

= −

The solution is 1, 3, 2x y z= = = − or using

ordered triplets (1, 3, 2)− .



52. 4

2 3 4 155 2 12

x y zx y zx y z

− + = −− + = −+ − =

Multiply the first equation by –3 and add to the second equation to eliminate y; add the first and third equations to eliminate y:

3 3 3 122 3 4 15x y zx y z

− + − =− + = −

33

x zz x

− + = −= −

45 2 12

6 8

x y zx y z

x z

− + = −+ − =

− =

Substitute and solve: 6 ( 3) 8

6 3 85 5

1

x xx x

xx

− − =− + =

==

3 1 3 2z x= − = − = −

12 5 2 12 5(1) 2( 2) 3y x z= − + = − + − = The solution is 1, 3, 2x y z= = = − or using

ordered triplets (1, 3, 2)− .

53. 2 3

2 4 72 2 3 4

x y zx y zx y z

+ − = −− + = −

− + − =

Add the first and second equations to eliminate z; multiply the second equation by 3 and add to the third equation to eliminate z: 2 32 4 7

3 2 10

x y zx y z

x y

+ − = −− + = −− = −

6 12 3 212 2 3 4

4 10 17

x y zx y z

x y

− + = −− + − =

− = −


15 10 504 10 17

x yx y

− + =− = −

11 333

xx

− == −

Substituting and solving for the other variables: 3( 3) 2 10

9 2 102 1

12

yyy

y

− − = −− − = −

− = −

=

13 2 3

23 1 3

1

1

z

z

z

z

− + − = −

− + − = −− = −

=

The solution is 1

3, , 12

x y z= − = = or using

ordered triplets 1

3, , 12

− .

54. 4 3 8

3 3 12

6 1

x y z

x y z

x y z

+ − = −− + =+ + =

Add the first and second equations to eliminate z; multiply the first equation by 2 and add to the third equation to eliminate z:

4 3 8

3 3 12

4 3 4

x y z

x y z

x y

+ − = −− + =+ =

2 8 6 16

6 1

3 9 15

x y z

x y z

x y

+ − = −+ + =+ = −

Multiply each side of the second result by 1/ 3− and add to the first result to eliminate y: 4 3 4

3 5

x y

x y

+ =− − =

3 9

3

x

x

==

Substituting and solving for the other variables: 3 3 5

3 8

83

y

y

y

+ = −= −

= −

83 6 1

32

6319

z

z

z

+ − + =

=

=

The solution is 8 1

3, ,3 9

x y z= = − = or using

ordered triplets 8 1

3, ,3 9

− .


1037


55. Let l be the length of the rectangle and w be the width of the rectangle. Then:

2l w= and 2 2 90l w+ =

Solve by substitution: 2(2 ) 2 90

4 2 90

6 90

15 feet

2(15) 30 feet

w w

w w

w

w

l

+ =+ =

=== =

The floor is 15 feet by 30 feet.

56. Let l be the length of the rectangle and w be the width of the rectangle. Then:

50l w= + and 2 2 3000l w+ =

Solve by substitution: 2( 50) 2 3000

2 100 2 3000

4 2900

725 meters

725 50 775 meters

w w

w w

w

w

l

+ + =+ + =

=== + =

The dimensions of the field are 775 meters by 725 meters.

57. Let x = the number of commercial launches and y = the number of noncommercial launches.

Then: 55x y+ = and 2 1y x= +

Solve by substitution: (2 1) 55

3 54

18

x x

x

x

+ + ===

2(18) 1

36 1

37

y

y

y

= += +=

In 2005 there were 18 commercial launches and 37 noncommercial launches.

58. Let x = the number of adult tickets sold and y = the number of senior tickets sold. Then:

325

9 7 2495

x y

x y

+ =+ =

Solve the first equation for y: 325y x= − Solve by substitution: 9 7(325 ) 2495

9 2275 7 2495

2 220

110

x x

x x

x

x

+ − =+ − =

==

325 110 215y = − = There were 110 adult tickets sold and 215 senior citizen tickets sold.

59. Let x = the number of pounds of cashews. Let y = is the number of pounds in the mixture. The value of the cashews is 5x . The value of the peanuts is 1.50(30) = 45. The value of the mixture is 3y . Then 30x y+ = represents the amount of mixture. 5 45 3x y+ = represents the value of the mixture.

Solve by substitution: 5 45 3( 30)

2 45

22.5

x x

x

x

+ = +==

So, 22.5 pounds of cashews should be used in the mixture.

60. Let x = the amount invested in AA bonds. Let y = the amount invested in the Bank Certificate.

a. Then 150,000x y+ = represents the total investment. 0.10 0.05 12,000x y+ = represents the earnings on the investment.

Solve by substitution: 0.10(150,000 ) 0.05 12,000

15,000 0.10 0.05 12,000

0.05 3000

60,000

y y

y y

y

y

− + =− + =

− = −=

150,000 60,000 90,000x = − = Thus, $90,000 should be invested in AA Bonds and $60,000 in a Bank Certificate.

b. Then 150,000x y+ = represents the total investment. 0.10 0.05 14,000x y+ = represents the earnings on the investment.

Solve by substitution: 0.10(150,000 ) 0.05 14,000

15,000 0.10 0.05 14,000

0.05 1000

20,000

y y

y y

y

y

− + =− + =

− = −=

150,000 20,000 130,000x = − = Thus, $130,000 should be invested in AA Bonds and $20,000 in a Bank Certificate.



61. Let x = the plane’s average airspeed and y = the average wind speed.

Rate Time Distance

With Wind 3 600

Against 4 600

x y

x y

+−

( )(3) 600

( )(4) 600

x y

x y

+ =− =

Multiply each side of the first equation by 13

,

multiply each side of the second equation by 14

,

and add the result to eliminate y 200

150

x y

x y

+ =− =

2 350

175xx

==

175 20025

yy

+ ==

The average airspeed of the plane is 175 mph, and the average wind speed is 25 mph.

62. Let x = the average wind speed and y = the distance.

Rate Time Distance

With Wind 150 2

Against 150 3

x y

x y

+−

(150 )(2)

(150 )(3)

x y

x y

+ =− =

Solve by substitution: (150 )(2) (150 )(3)

300 2 450 3

5 150

30

x x

x x

x

x

+ = −+ = −

==

Thus, the average wind speed is 30 mph.

63. Let x = the number of $25-design. Let y = the number of $45-design. Then x y+ = the total number of sets of dishes. 25 45x y+ = the cost of the dishes.

Setting up the equations and solving by substitution:

200

25 45 7400

x y

x y

+ =+ =

Solve the first equation for y, the solve by substitution: 200y x= −

25 45(200 ) 7400

25 9000 45 7400

20 1600

80

x x

x x

x

x

+ − =+ − =

− = −=

200 80 120y = − = Thus, 80 sets of the $25 dishes and 120 sets of the $45 dishes should be ordered.

64. Let x = the cost of a hot dog. Let y = the cost of a soft drink. Setting up the equations and solving by substitution:

10 5 35.00

7 4 25.25

x y

x y

+ =+ =

10 5 35.00

2 7

7 2

x y

x y

y x

+ =+ =

= −

7 4(7 2 ) 25.25

7 28 8 25.25

2.75

2.75

x x

x x

x

x

+ − =+ − =

− = −=

7 2(2.75) 1.50y = − = A single hot dog costs $2.75 and a single soft drink costs $1.50.

65. Let x = the cost per package of bacon. Let y = the cost of a carton of eggs. Set up a system of equations for the problem:

3 2 13.45

2 3 11.45

x y

x y

+ =+ =

Multiply each side of the first equation by 3 and each side of the second equation by –2 and solve by elimination:

9 6 40.35

4 6 22.90

x y

x y

+ =− − = −

5 17.45

3.49

x

x

==

Substitute and solve for y: 3(3.49) 2 13.45

10.47 2 13.45

2 2.98

1.49

y

y

y

y

+ =+ =

==

A package of bacon costs $3.49 and a carton of eggs cost $1.49. The refund for 2 packages of bacon and 2 cartons of eggs will be 2($3.49) + 2($1.49) = $9.96.


1039


66. Let x = Pamela’s average speed in still water. Let y = the speed of the current.

Rate Time Distance

Downstream 3 15

Upstream 5 15

x y

x y

+−

Set up a system of equations for the problem:

3( ) 15

5( ) 15

x y

x y

+ =− =

Multiply each side of the first equation by 13

,

multiply each side of the second equation by 15

,

and add the result to eliminate y:

5

3

x y

x y

+ =− =

2 8

4

x

x

==

4 5

1

y

y

+ ==

Pamela's average speed is 4 miles per hour and the speed of the current is 1 mile per hour.

67. Let x = the # of mg of compound 1. Let y = the # of mg of compound 2. Setting up the equations and solving by substitution:

0.2 0.4 40 vitamin C

0.3 0.2 30 vitamin D

x y

x y

+ =+ =

Multiplying each equation by 10 yields 2 4 400

6 4 600

x y

x y

+ =+ =

Subtracting the bottom equation from the top equation yields

( )2 4 6 4 400 600

2 6 200

4 200

50

x y x y

x x

x

x

+ − + = −− = −− = −

=

( )2 50 4 400

100 4 400

4 300

30075

4

y

y

y

y

+ =+ =

=

= =

So 50 mg of compound 1 should be mixed with 75 mg of compound 2.

68. Let x = the # of units of powder 1. Let y = the # of units of powder 2. Setting up the equations and solving by substitution:

120.2 0.4 12 vitamin B

0.3 0.2 12 vitamin E

x y

x y

+ =+ =

Multiplying each equation by 10 yields

2 4 120

6 4 240

x y

x y

+ =+ =

Subtracting the bottom equation from the top equation yields

( )2 4 6 4 120 240

4 120

30

x y x y

x

x

+ − + = −− = −

=

( )2 30 4 120

60 4 120

4 60

6015

4

y

y

y

y

+ =+ =

=

= =

So 30 units of powder 1 should be mixed with 15 units of powder 2.

69. 2y ax bx c= + + At (–1, 4) the equation becomes:

24 (–1) ( 1)

4

a b c

a b c

= + − += − +

At (2, 3) the equation becomes: 23 (2) (2)

3 4 2

a b c

a b c

= + += + +


1

a b c

c

= + +=

The system of equations is: 4

4 2 3

1

a b c

a b c

c

− + =+ + =

=

Substitute 1c = into the first and second equations and simplify:

1 4

3

3

a b

a b

a b

− + =− =

= +

4 2 1 3

4 2 2

a b

a b

+ + =+ =

Solve the first result for a, substitute into the second result and solve:



4( 3) 2 2

4 12 2 2

6 10

53

b b

b b

b

b

+ + =+ + =

= −

= −

5 43

3 3a = − + =

The solution is 4 5

, , 13 3

a b c= = − = . The

equation is 24 51

3 3y x x= − + .

70. 2y ax bx c= + + At (–1, –2) the equation becomes:

22 ( 1) ( 1)

2

a b c

a b c

− = − + − +− + = −

At (1, –4) the equation becomes: 24 (1) (1)

4

a b c

a b c

− = + ++ + = −


4 2 4

a b c

a b c

= + ++ + =


– 4

4 2 4

a b c

a b c

a b c

− + = −+ + =+ + =

Multiply the first equation by –1 and add to the second equation; multiply the first equation by –1 and add to the third equation to eliminate c:

2

– 4

a b c

a b c

− + − =+ + =

2

4 2 4

a b c

a b c

− + − =+ + =

2 2

1

b

b

= −= −

3 3 6

2

a b

a b

+ =+ =

Substitute and solve: ( 1) 2

3

a

a

+ − ==

4

3 ( 1) 4

6

c a b= − − −= − − − −= −

The solution is 3, 1, 6a b c= = − = − . The

equation is 23 6y x x= − −

71. 0.06 5000 240

0.06 6000 900

Y r

Y r

− =+ =

Multiply the first equation by 1− , the add the result to the second equation to eliminate Y.

0.06 5000 240

0.06 6000 900

Y r

Y r

− + = −+ =

11000 660

0.06

r

r

==

Substitute this result into the first equation to find Y. 0.06 5000(0.06) 240

0.06 300 240

0.06 540

9000

Y

Y

Y

Y

− =− =

==

The equilibrium level of income and interest rates is $9000 million and 6%.

72. 0.05 1000 10

0.05 800 100

Y r

Y r

− =+ =

Multiply the first equation by 1− , the add the result to the second equation to eliminate Y.

0.05 1000 10

0.05 800 100

Y r

Y r

− + = −+ =

1800 90

0.05

r

r

==

Substitute this result into the first equation to find Y. 0.05 1000(0.05) 10

0.05 50 10

0.05 60

1200

Y

Y

Y

Y

− =− =

==

The equilibrium level of income and interest rates is $1200 million and 5%.

73. 2 1 3

1 2

2 3

5 3 5 0

10 5 7 0

I I I

I I

I I

= +− − =− − =

Substitute the expression for 2I into the second and third equations and simplify:

1 1 3

1 3

5 3 5( ) 0

8 5 5

I I I

I I

− − + =− − = −

1 3 3

1 3

10 5( ) 7 0

5 12 10

I I I

I I

− + − =− − = −

Multiply both sides of the first result by 5 and multiply both sides of the second result by –8 to eliminate 1I :


1041


1 3

1 3

40 25 25

40 96 80

I I

I I

− − = −+ =

3

3

71 55

5571

I

I

=

=


1

1

1

1

558 5 5

71275

8 571

808

711071

I

I

I

I

− − = −

− − = −

− = −

=

210 55 6571 71 71

I = + =

The solution is 1 2 310 65 55

, ,71 71 71

I I I= = = .

74. 3 1 2

3 2

1 2

8 4 6

8 4 6

I I I

I I

I I

= += += +

Substitute the expression for 3I into the second equation and simplify:

1 2 2

1 2

1 2

8 4( ) 6

8 4 10

4 10 8

I I I

I I

I I

= + += +

+ =

1 2

1 2

8 4 6

8 6 4

I I

I I

= +− =

Multiply both sides of the first result by –2 and add to the second result to eliminate 1I :

1 2

1 2

8 20 16

8 6 4

I I

I I

− − = −− =

2

2

26 12

12 626 13

I

I

− = −−= =−


1

1

1

1

64 10 8

1360

4 813

444

131113

I

I

I

I

+ =

+ =

=

=

3 1 211 6 1713 13 13

I I I= + = + =

The solution is 1 2 311 6 17

, ,13 13 13

I I I= = = .

75. Let x = the number of orchestra seats. Let y = the number of main seats. Let z = the number of balcony seats. Since the total number of seats is 500,

500x y z+ + = . Since the total revenue is $17,100 if all seats are sold, 50 35 25 17,100x y z+ + = . If only half of the orchestra seats are sold, the revenue is $14,600.

So, 1

50 35 25 14,6002

x y z+ + = .

Thus, we have the following system: 500

50 35 25 17,100

25 35 25 14,600

x y z

x y z

x y z

+ + =+ + =+ + =

Multiply each side of the first equation by –25 and add to the second equation to eliminate z; multiply each side of the third equation by –1 and add to the second equation to eliminate z:

25 25 25 12,500

50 35 25 17,100

25 10 4600

x y z

x y z

x y

− − − = −+ + =+ =

50 35 25 17,100

25 35 25 14,600

x y z

x y z

+ + =− − − = −

25 2500

100

x

x

==

Substituting and solving for the other variables: 25(100) 10 4600

2500 10 4600

10 2100

210

y

y

y

y

+ =+ =

==

100 210 500

310 500

190

z

z

z

+ + =+ =

=

There are 100 orchestra seats, 210 main seats, and 190 balcony seats.



76. Let x = the number of adult tickets. Let y = the number of child tickets. Let z = the number of senior citizen tickets. Since the total number of tickets is 405,

405x y z+ + = . Since the total revenue is $2320, 8 4.50 6 2320x y z+ + = . Twice as many children's tickets as adult tickets are sold. So, 2y x= . Thus, we have the following system:

405

8 4.50 6 2320

2

x y z

x y z

y x

+ + =+ + =

=

Substitute for y in the first two equations and simplify:

(2 ) 405

3 405

x x z

x z

+ + =+ =

8 4.50(2 ) 6 2320

17 6 2320

x x z

x z

+ + =+ =

Multiply the first result by –6 and add to the second result to eliminate z:

18 6 2430

17 6 2320

x z

x z

− − = −+ =

110

110

x

x

− = −=

2

2(110)

220

y x===

3 405

3(110) 405

330 405

75

x z

z

z

z

+ =+ =+ =

=

There were 110 adults, 220 children, and 75 senior citizens that bought tickets.

77. Let x = the number of servings of chicken. Let y = the number of servings of corn. Let z = the number of servings of 2% milk.

Protein equation: 30 3 9 66x y z+ + = Carbohydrate equation: 35 16 13 94.5x y z+ + = Calcium equation: 200 10 300 910x y z+ + =

Multiply each side of the first equation by –16 and multiply each side of the second equation by 3 and add them to eliminate y; multiply each side of the second equation by –5 and multiply each side of the third equation by 8 and add to eliminate y:

480 48 144 1056

105 48 39 283.5

375 105 772.5

x y z

x y z

x z

− − − = −+ + =

− − = −

175 80 65 472.5

1600 80 2400 7280

1425 2335 6807.5

x y z

x y z

x z

− − − = −+ + =

+ =

Multiply each side of the first result by 19 and multiply each side of the second result by 5 to eliminate x:

7125 1995 14,677.5

7125 11,675 34,037.5

x z

x z

− − = −+ =

9680 19,360

2

z

z

==

Substituting and solving for the other variables: 375 105(2) 772.5

375 210 772.5

375 562.5

1.5

x

x

x

x

− − = −− − = −

− = −=

30(1.5) 3 9(2) 66

45 3 18 66

3 3

1

y

y

y

y

+ + =+ + =

==

The dietitian should serve 1.5 servings of chicken, 1 serving of corn, and 2 servings of 2% milk.

78. Let x = the amount in Treasury bills. Let y = the amount in Treasury bonds. Let z = the amount in corporate bonds.

Since the total investment is $20,000, 20,000x y z+ + =

Since the total income is to be $1390, 0.05 0.07 0.10 1390x y z+ + =

The investment in Treasury bills is to be $3000 more than the investment in corporate bonds. So, 3000x z= +

Substitute for x in the first two equations and simplify: (3000 ) 20,000

2 17,000

z y z

y z

+ + + =+ =

5(3000 ) 7 10 139,000

7 15 124,000

z y z

y z

+ + + =+ =


1043



7 14 119,000

7 15 124,000

5,000

y z

y z

z

− − = −+ =

=

3000 3000 5000 8000x z= + = + = 2 17,000

2(5000) 17,000

10,000 17,000

7000

y z

y

y

y

+ =+ =+ =

=

Kelly should invest $8000 in Treasury bills, $7000 in Treasury bonds, and $5000 in corporate bonds.

79. Let x = the price of 1 hamburger. Let y = the price of 1 order of fries. Let z = the price of 1 drink.

We can construct the system 8 6 6 26.10

10 6 8 31.60

x y z

x y z

+ + =+ + =

A system involving only 2 equations that contain 3 or more unknowns cannot be solved uniquely.

Multiply the first equation by 12

− and the

second equation by 12

, then add to eliminate y:

4 3 3 13.05

5 3 4 15.80

x y z

x y z

− − − = −+ + =

2.75

2.75

x z

x z

+ == −

Substitute and solve for y in terms of z: ( )5 2.75 3 4 15.80

13.75 3 15.80

3 2.05

1 413 60

z y z

y z

y z

y z

− + + =+ − =

= +

= +

Solutions of the system are: 2.75x z= − , 1 413 60

y z= + .

Since we are given that 0.60 0.90z≤ ≤ , we choose values of z that give two-decimal-place values of x and y with 1.75 2.25x≤ ≤ and 0.75 1.00y≤ ≤ . The possible values of x, y, and z are shown in the table.

x y z

2.13 0.89 0.62

2.10 0.90 0.65

2.07 0.91 0.68

2.04 0.92 0.71

2.01 0.93 0.74

1.98 0.94 0.77

1.95 0.95 0.80

1.92 0.96 0.83

1.89 0.97 0.86

1.86 0.98 0.89 80. Let x = the price of 1 hamburger.

Let y = the price of 1 order of fries. Let z = the price of 1 drink

We can construct the system 8 6 6 26.1010 6 8 31.603 2 4 10.95

x y zx y zx y z

+ + =+ + =+ + =

Subtract the second equation from the first equation to eliminate y: 8 6 6 26.1010 6 8 31.60

2 2 5.5

x y zx y z

x z

+ + =+ + =− − = −

Multiply the third equation by –3 and add it to the second equation to eliminate y: 10 6 8 31.609 6 12 32.85

4 1.25

x y zx y z

x z

+ + =− − − = −

− = −

Multiply the second result by 2 and add it to the first result to eliminate x:

2 2 5.52 8 2.5x zx z

− − = −− = −

10 80.8

zz

− = −=

Substitute for z to find the other variables: 4(0.8) 1.25

3.2 1.251.95

xx

x

− = −− = −

=

3(1.95) 2 4(0.8) 10.955.85 2 3.2 1.095

2 1.90.95

yy

yy

+ + =+ + =

==

Therefore, one hamburger costs $1.95, one order of fries costs $0.95, and one drink costs $0.80.



81. Let x = Beth’s time working alone. Let y = Bill’s time working alone. Let z = Edie’s time working alone.

We can use the following tables to organize our work:

Beth Bill Edie

Hours to do job

Part of job done 1 1 1in 1 hour

x y z

x y z

In 10 hours they complete 1 entire job, so 1 1 1

10 1

1 1 1 110

x y z

x y z

+ + =

+ + =

Bill Edie

Hours to do job

Part of job done 1 1in 1 hour

y z

y z

In 15 hours they complete 1 entire job, so 1 1

15 1

1 1 115

y z

y z

+ =

+ =

.

Beth Bill Edie

Hours to do job

Part of job done 1 1 1in 1 hour

x y z

x y z

With all 3 working for 4 hours and Beth and Bill working for an additional 8 hours, they complete

1 entire job, so 1 1 1 1 1

4 8 1

12 12 41

x y z x y

x y z

+ + + + =

+ + =

We have the system 1 1 1 1

10

1 1 115

12 12 41

x y z

y z

x y z

+ + =

+ =

+ + =

Subtract the second equation from the first equation:

1 1 1 1

10

1 1 1

15

x y z

y z

+ + =

+ =

1 1

3030

xx

=

=

Substitute x = 30 into the third equation: 12 12 4

130

12 4 35

y z

y z

+ + =

+ =.

Now consider the system consisting of the last result and the second original equation. Multiply the second original equation by –12 and add it to the last result to eliminate y:

12 12 1215

12 4 3

5

y z

y z

− − −+ =

+ =

8 3

1540

zz

− = −

=

Plugging z = 40 to find y: 12 4 3

5

12 4 340 5

12 12

24

y z

y

y

y

+ =

+ =

=

=

Working alone, it would take Beth 30 hours, Bill 24 hours, and Edie 40 hours to complete the job.

82. – 84. Answers will vary.

Section 11.2: Systems of Linear Equations: Matrices

1045


Section 11.2

1. matrix

2. augmented

3. third; fifth

4. True

5. Writing the augmented matrix for the system of equations:

5 5 5 514 3 6 3 64

x y

x y

− = −→+ =


3 4 7 3 744 2 5 54 2

x y

x y

+ =→

− = −

7. 2 3 6 0

4 6 2 0

x y

x y

+ − =− + =

Write the system in standard form and then write the augmented matrix for the system of equations:

2 3 6 3 624 6 2 4 6 2

x y

x y

+ =→

− = − − −

8. 9 0

3 4 0

x y

x y

− =− − =


9 0 9 013 4 3 1 4

x y

x y

− = −→− = −


0.01 0.03 0.06 0.01 0.060.030.13 0.10 0.20 0.13 0.10 0.20

x y

x y

− = −→+ =


4 3 3 4 3 33 2 4 3 2 41 1 2 1 214 3 3 3 34

x y

x y

− = −→

− + = −


10 1 1 1 10

3 3 5 3 3 0 5

2 2 1 1 2 2

x y z

x y

x y z

− + = −+ = →

+ + =


5 0 5 1 1 0

5 1 1 0 5

2 3 2 2 0 3 2

x y z

x y

x z

− − = − −+ = →

− = −


2 1 1 1 2

3 2 2 3 2 0 2

5 3 1 5 3 1 1

x y z

x y

x y z

+ − = −− = → −+ − = −

14. 2 3 4 0

5 2 0

2 3 2

x y z

x z

x y z

+ − =− + =+ − = −


2 3 4 0 2 3 4 0

5 2 1 0 5 2

2 3 2 1 2 3 2

x y z

x z

x y z

+ − = −− = − → − −

+ − = − − −


10 1 1 1 10

2 2 1 2 1 2 1

3 4 5 3 4 0 5

4 5 0 4 5 1 0

x y z

x y z

x y

x y z

− − = − −+ + = − −

→− + = −− + = −


2 5 1 1 2 1 5

3 4 2 2 1 3 4 2 2

3 5 1 3 1 5 1 1

x y z w

x y z w

x y z w

− + − = − −+ − + = → −

− − − = − − − − −



17. 1 3 2 3 2

2 5 5 2 5 5

x y

x y

− − − = −→

− − =

2 1 22R r r= − +

1 3 2 1 3 2

2 5 5 2(1) 2 2( 3) 5 2( 2) 5

1 3 2

0 1 9

− − − −→

− − + − − − − +

− −→

18. 1 3 3 3 3

2 5 4 2 5 4

x y

x y

− − − = −→

− − − = −

2 1 22R r r= − +

1 3 3 1 3 3

2 5 4 2(1) 2 2( 3) 5 2( 3) 4

1 3 3

0 1 2

− − − −→

− − − + − − − − − −

− −→

19. 1 3 4 3 3 4 3

3 5 6 6 3 5 6 6

5 3 4 6 5 3 4 6

x y z

x y z

x y z

− − + =− → − + =

− − + + =

2 1 23R r r= − +

1 3 4 33 5 6 65 3 4 6

1 3 4 33(1) 3 3( 3) 5 3(4) 6 3(3) 6

5 3 4 6

1 3 4 30 4 6 35 3 4 6

−−

−−

→ − + − − − − + − +−

−→ − −

−

3 1 35R r r= +

1 3 4 30 4 6 35 3 4 6

1 3 4 30 4 6 3

5(1) 5 5( 3) 3 5(4) 4 5(3) 6

1 3 4 30 4 6 30 12 24 21

−− −

−−

→ − −− − + + +−

→ − −−

20. 1 3 3 5 3 3 5

4 5 3 5 4 5 3 5

3 2 4 6 3 2 4 6

x y z

x y z

x y z

− − − + = −− − − − → − − − = −− − − − + =

2 1 24R r r= +

1 3 3 54 5 3 53 2 4 6

1 3 3 54(1) 4 4( 3) 5 4(3) 3 4( 5) 5

3 2 4 6

1 3 3 50 17 9 253 2 4 6

− −− − − −− −

− −→ − − − − − −

− −− −

→ − −− −

3 1 33R r r= +

1 3 3 54 5 3 53 2 4 6

1 3 3 50 17 9 25

3(1) 3 3( 3) 2 3(3) 4 3( 5) 6

1 3 3 50 17 9 250 11 13 9

− −− − − −− −

− −→ − −

− − − + − +− −

→ − −− −

21. 1 3 2 6 3 2 6

2 5 3 4 2 5 3 4

3 6 4 6 3 6 4 6

x y z

x y z

x y z

− − − + = −− − → − + = −

− − − − + =

2 1 22R r r= − +

1 3 2 62 5 3 43 6 4 6

1 3 2 62(1) 2 2( 3) 5 2(2) 3 2( 6) 4

3 6 4 6

1 3 2 60 1 1 83 6 4 6

− −− −

− −− −

→ − + − − − − + − − −− −

− −→ −

− −


1047


3 1 33R r r= +

1 3 2 60 1 1 83 6 4 6

1 3 2 60 1 1 8

3(1) 3 3( 3) 6 3(2) 4 3( 6) 6

1 3 2 60 1 1 80 15 10 12

− −−

− −− −

→ −− − − + − +− −

→ −− −

22. 1 3 4 6 3 4 6

6 5 6 6 6 5 6 6

1 1 4 6 4 6

x y z

x y z

x y z

− − − − − = −− − → − + = −

− − + + =

a. 2 1 26R r r= − +

1 3 4 66 5 6 61 1 4 6

1 3 4 66(1) 6 6( 3) 5 6( 4) 6 6( 6) 6

1 1 4 6

1 3 4 60 13 30 301 1 4 6

− − −− −

−− − −

→ − + − − − − − + − − −−

− − −→

−

b. 3 1 3R r r= +

1 3 4 6 1 3 4 66 5 6 6 6 5 6 61 1 4 6 1 1 3 1 4 4 6 6

1 3 4 66 5 6 60 2 0 0

− − − − − −− − → − −

− − − + − + − +− − −

→ − −−

23. 5 3 1 2 5 3 2

2 5 6 2 2 5 6 2

4 1 4 6 4 4 6

x y z

x y z

x y z

− − − + = −− − → − + = −

− − + + =

1 2 12R r r= − +

5 3 1 22 5 6 24 1 4 6

2(2) 5 2( 5) 3 2(6) 1 2( 2) 22 5 6 24 1 4 6

1 7 11 22 5 6 24 1 4 6

− −− −

−− + − − − − + − − −

→ − −−

−→ − −

−

3 2 32R r r= +

1 7 11 22 5 6 24 1 4 6

1 7 11 22 5 6 2

2(2) ( 4) 2( 5) 1 2(6) 4 2( 2) 6

1 7 11 22 5 6 20 9 16 2

−− −

−−

→ − −+ − − + + − +

−→ − −

−

24. 4 3 1 2 4 3 23 5 2 6 3 5 2 63 6 4 6 3 6 4 6

x y zx y zx y z

− − − − =− → − + =

− − − − + =

1 2 1R r r= − +

4 3 1 23 5 2 63 6 4 6

(3) 4 ( 5) 3 (2) 1 (6) 23 5 2 63 6 4 6

1 2 3 43 5 2 63 6 4 6

− −−

− −− + − − − − − − +

→ −− −

− −→ −

− −

3 2 3R r r= +

1 2 3 43 5 2 63 6 4 6

1 2 3 43 5 2 6

3 ( 3) 5 ( 6) 2 4 6 6

1 2 3 43 5 2 60 11 6 12

− −−

− −− −

→ −+ − − + − + +

− −→ −

−

25. 5

1

x

y

== −

Consistent; 5, 1,x y= = − or using ordered pairs

(5, 1)− .

26. 4

0

x

y

= −=

Consistent; 4, 0,x y= − = or using ordered pairs

( 4, 0)− .



27. 1

2

0 3

x

y

===

Inconsistent

28. 0

0

0 2

x

y

===

Inconsistent

29. 2 1

4 2

0 0

x z

y z

+ = −− = −

=

Consistent; 1 2

2 4

is any real number

x z

y z

z

= − −= − +

or {( , , ) | 1 2 , 2 4 ,x y z x z y z= − − = − + z is any real number}

30. 4 4

3 2

0 0

x z

y z

+ =+ =

=

Consistent; 4 4

2 3

is any real number

x z

y z

z

= −= −

or {( , , ) | 4 4 , 2 3 ,x y z x z y z= − = − z is any real number}

31. 1

2 4

3 4

1

2

2 3

x

x x

x x

=+ =

+ =

Consistent;

1

2 4

3 4

4

1

2

3 2

is any real number

x

x x

x x

x

== −= −

or 1 2 3 4 1 2 4{( , , , ) | 1, 2 ,x x x x x x x= = −

3 4 43 2 , is any real number}x x x= −

32. 1

2 4

3 4

1

2 2

3 0

x

x x

x x

=+ =+ =

Consistent;

1

2 4

3 4

4

1

2 2

3

is any real number

x

x x

x x

x

== −= −

or 1 2 3 4 1 2 4 3 4{( , , , ) | 1, 2 2 , 3 ,x x x x x x x x x= = − = −

4 is any real number}x

33. 1 4

2 3 4

4 2

3 3

0 0

x x

x x x

+ =+ + =

=

Consistent;

1 4

2 3 4

3 4

2 4

3 3

, are any real numbers

x x

x x x

x x

= −= − −

or 1 2 3 4 1 4 2 3 4{( , , , ) | 2 4 , 3 3 ,x x x x x x x x x= − = − −

3 4and are any real numbers}x x

34. 1

2

3 4

1

2

2 3

x

x

x x

==

+ =

Consistent;

1

2

3 4

4

1

2

3 2

is any real number

x

x

x x

x

=== −

or 1 2 3 4 1 2 3 4{( , , , ) | 1, 2, 3 2 ,x x x x x x x x= = = −

4 is any real number}x

35.

1 4

2 4

3 4

2

2 2

0

0 0

x x

x x

x x

+ = −+ =

− ==

Consistent;

1 4

2 4

3 4

4

2

2 2

is any real number

x x

x x

x x

x

= − −= −=

or 1 2 3 4 1 4 2 4{( , , , ) | 2 , 2 2 ,x x x x x x x x= − − = −

3 4 4, is any real number}x x x=


1049


36.

1

2

3

4

1

2

3

0

x

x

x

x

====

Consistent;

1

2

3

4

1

2

3

0

x

x

x

x

====

or (1,2,3,0)

37. 8

4

x y

x y

+ =− =

Write the augmented matrix:

( )

( )

( )

2 1 2

12 22

1 2 1

8 81 1 1 1 01 1 4 2 4

81 1 0 1 2

0 61 0 1 2

R r r

R r

R r r

→ = − +− − −

→ = −

→ = − +

The solution is 6, 2x y= = or using ordered pairs (6, 2).

38. 2 5

3

x y

x y

+ =+ =


( )

( )

( )

2 1 2

2 2

1 2 1

5 51 2 1 2 3 01 1 1 2

51 2 0 1 2

01 1 20 1 2

R r r

R r

R r r

→ = − +− −

→ = −

→ = − +

The solution is 1, 2x y= = or using ordered pairs (1, 2).

39. 2 4 2

3 2 3

x y

x y

− = −+ =


( )

( )

( )

( )

11 12

2 1 2

12 283

4

12

1 2 134

2 1 14 2 2 3 3 3 32 2

1 12 3 0 8 6

1 12 0 1

01 2 0 1

R r

R r r

R r

R r r

−− − − =→

−− = − +→

−− =→

= +→

The solution is 1 3

,2 4

x y= = or using ordered pairs

1 3,

2 4.

40. 3 3 3

84 2

3

x y

x y

+ =

+ =


( )11 18 38

3 3

3 3 3 1 1 1

4 2 4 2R r=→

( )

( )

( )

2 1 243

12 22

23

13 1 2 123

1 1 1 4 0 2

1 1 1

0 1

01 0 1

R r r

R r

R r r

= − +→−−

= −→

= − +→

The solution is 1 2

,3 3

x y= = or using ordered

pairs 1 2

,3 3

.

41. 2 4

2 4 8

x y

x y

+ =+ =


( )2 1 21 2 4 1 2 4 2

8 0 0 02 4R r r= − +→

This is a dependent system. 2 4

4 2

x y

x y

+ == −

The solution is 4 2 , is any real numberx y y= −

or {( , ) | 4 2 , is any real number}x y x y y= −



42. 3 7

9 3 21

x y

x y

− =− =


( )

( )

713

1 13

713 3

2 1 2

113 71 3

9 3 21 9 3 21

-1 90 0 0

R r

R r r

−− → =− −

= − +→

This is a dependent system. 3 7

3 7

x y

x y

− =− =

The solution is 3 7, is any real numbery x x= −

or {( , ) | 3 7, is any real number}x y y x x= −

43. 2 3 6

12

x y

x y

+ =

− =


( )

( )

( )

( )

312

1 121 12 2

32

2 1 25 52 2

322 2 25

332 1 2 12

3 62 31 1 1 1 1

31 0

31 0 1 1

01 0 1 1

R r

R r r

R r

R r r

→ =− −

→ = − +− −

→ = −

→ = − +

The solution is 3

, 12

x y= = or 3

, 12

.

44. 1

22

2 8

x y

x y

+ = −

− =


( )

( )

( )

( )

12 1 1

2 1 2

12 24

1 2 1

1 2 41 2 281 281 2

1 2 4 0 124

1 2 4 0 31

01 2 20 31

R r

R r r

R r

R r r

−− → =−−

−→ = − +−

−→ = −−

→ = − +−

The solution is 2, 3x y= = − or (2, 3)− .

45. 3 5 3

15 5 21

x y

x y

− =+ =


( )

( )

( )

( )

513 1 13

53

2 1 2

53 1

2 23015

453

1 2 1315

3 5 3 1 1 15 5 21 15 5 21

1 1 150 30 6

1 1 0 1

01 0 1

R r

R r r

R r

R r r

− −→ =

−→ = − +

−→ =

→ = +

The solution is 4 1

,3 5

x y= = or 4 1

,3 5

.

46. 2 1

1 32 2

x y

x y

− = −

+ =

( )

( )

( )

1 112 2

1 1231 312 2 2 2

1 12 2 2 1 2

112 1 2 12

2 1 1 1 1 1

1 10 1 2

01 0 1 2

R r

R r r

R r r

− − − −→ =

− −→ = − +

→ = +

The solution is 1

, 22

x y= = or 1

, 22

.

47. 6

2 3 16

2 4

x y

x z

y z

− =− =+ =


( )

( )

2 1 2

3 12 22 2

0 61 1

0 3 162

0 2 1 4

0 61 1

0 3 22 4

0 2 1 4

0 61 1

0 1 2

0 2 1 4

R r r

R r

−−

−→ − = − +

−−→ =


1051


( )

32

1 2 132

3 2 3

323 1

3 32 4

31 3 12

32 3 22

0 81

0 1 2 2

0 0 04

0 81

0 1 2

0 0 01

0 0 81

0 0 1 2

0 0 01

R r r

R r r

R r

R r r

R r r

−= +

−→= − +

−

−→ =

= +→

= +

The solution is 8, 2, 0x y z= = = or (8, 2, 0).

48. 2 4

2 4 0

3 2 11

x y

y z

x z

+ = −− + =

− = −


( )12

11 12

02 1 40 0423 0 112

201

0 0 423 0 112

R r

−−

−−

−→ =−

−−

( )

( )

( )

12

3 1 3

32

12

12 22

32

11 2 12

33 2 32

13 35

201

0 0 342

0 52

201

0 0 1 2

0 52

201 1

0 0 1 20 0 5 5

201 1

0 0 1 20 0 1 1

R r r

R r

R r r

R r r

R r

−→ = − +−

− −−

−→ = −−

− −−

− = − +→ −

= +− −

−→ = −−

1 3 1

2 3 2

0 0 31

0 0 1 22

0 0 1 1

R r r

R r r

−= − +

→= +

The solution is 3, 2, 1x y z= − = = or ( 3, 2,1)− .

49. 2 3 7

2 4

3 2 2 10

x y z

x y z

x y z

− + =+ + =

− + − = −


( )

2 1 2

3 1 3

12 25

1 2 1

3 2 3

3 71 22 1 1 4

3 102 2

3 71 22

0 5 5 10 3

0 7 114

3 71 20 1 1 20 7 114

0 31 12

0 1 1 24

0 0 3 3

R r r

R r r

R r

R r r

R r r

−

− −−

− = − +→ − −

= +−

−→ =− −

−

= +→ − −

= +

( )13 33

0 31 1

0 1 1 20 0 1 1

R r→ =− −

1 3 1

2 3 2

0 01 2

0 0 1 1

0 0 1 1

R r r

R r r

= − +→ −

= +

The solution is 2, 1, 1x y z= = − = or (2, 1,1)− .

50. 2 3 0

2 2 7

3 4 3 7

x y z

x y z

x y z

+ − =− + + = −

− − =


( )31

2 21

1 12

312 2

2 1 2

3 1 33112 2

2 1 3 0

2 2 1 7

3 4 3 7

1 0

2 2 1 7

3 4 3 7

1 02

0 3 2 7 3

0 7

R r

R r r

R r r

−− −

− −

−→ − − =

− −

−= +

→ − −= − +

−



( )31

2 272 1

2 23 3 3311

2 2

1 0

0 1

0 7

R r

−→ − − =

−

( )

7 76 6 1

1 2 12723 3 11

3 2 3213 356 6

7 76 6

7 623 33 3 13

3513

5613 7

1 3 16713 2

2 3 235 313

1 0

0 1

0 0

1 0

0 1

0 0 1

1 0 0

0 1 0

0 0 1

R r r

R r r

R r

R r r

R r r

−= − +

→ − −= +

− −

−→ − − = −

= +→ −

= +


, ,13 13 13

x y z= = − = or

56 7 35, ,

13 13 13− .

51. 2 2 2 2

2 3 2

3 2 0

x y z

x y z

x y

− − =+ + =+ =


( )

( )

11 12

2 1 2

3 1 3

3 2 3

2 22 2

32 1 2

3 0 02

1 1 1 1

32 1 2

3 0 02

1 1 1 12

0 5 3 0 3

0 5 3 3

1 1 1 1

0 5 3 0

0 0 0 3

R r

R r r

R r r

R r r

− −

− −→ =

− −= − +

→= − +

−

− −→ = − +

−

There is no solution. The system is inconsistent.

52. 2 3 0

2 5

3 4 1

x y z

x y z

x y z

− − =− + + =

− − =


( )

1 2

2 1 2

3 1 3

3 2 3

3 02 151 2 1

3 1 14

51 12Interchange

3 02 1 and

3 1 14

51 122

0 10 1 13

0 162 2

51 120 10 21 10 0 0 4

r r

R r rR r r

R r r

− −−

−−

−−−−→ −

−−−

−−− = − +→

= − +

−−−→ = − +

−


53. 1

2 3 4

3 2 7 0

x y z

x y z

x y z

− + + = −− + − = −

− − =


( )1 1

2 1 2

3 1 3

1 2 1

3 2 3

1 1 1 1431 2

23 7 0

1 1 1 1431 2

23 7 0

1 1 1 140 3 1

340 31

20 5140 3 1

0 0 0 0

R r

R r rR r r

R r rR r r

− −−−−

− −

− −−−→ − = −

− −

− − = +−→ −

= − +− −

−−= +

−→ −= − +

The matrix in the last step represents the system 5 2

4 3

0 0

x z

y z

− = −− = −

= or, equivalently,

5 2

4 3

0 0

x z

y z

= −= −=

The solution is 5 2x z= − , 4 3y z= − , z is any

real number or {( , , ) | 5 2, 4 3,x y z x z y z= − = − z is any real number}.


1053


54. 2 3 0

3 2 2 2

5 3 2

x y z

x y z

x y z

− − =+ + =+ + =


1 3

2 1 2

3 1 3

3 2 37 413 13 1

2 213

3 02 1

3 2 2 2

5 31 2

5 31 2Interchange

3 2 2 2 and

3 02 1

5 31 23

0 13 7 42

0 13 7 4

5 31 2

0 1

0 0 0 0

r r

R r r

R r r

R r r

R r

− −

→− −

= − +→ − − −

= − +− − −

= − +→

= −

( )

6413 13

7 41 2 113 13

01

0 51

0 0 0 0

R r r→ = − +


13 137 4

13 130 0

x z

y z

+ =

+ =

=

or, equivalently, 4 6

13 137 4

13 130 0

x z

y z

= − +

= − +

=

The solution is 4 6

13 13x z= − + ,

7 413 13

y z= − + ,

z is any real number or 4 6

( , , ) ,13 13

x y z x z= − +


13 13y z z= − + .

55. 2 2 3 6

4 3 2 0

2 3 7 1

x y z

x y z

x y z

− + =− + =

− + − =


2 3 62

3 04 22 3 7 1

−−

− −

( )32

11 12

32

2 1 2

3 1 3

52

1 2 1

3 2 3

31 1

3 04 2 2 3 7 1

31 14

40 1 122

40 71

0 9140 1 12

0 0 0 19

R r

R r r

R r r

R r r

R r r

−−→ =

− −

−= − +

−→ −= +

−

− −= +

−→ −= − +


56. 3 2 2 6

7 3 2 1

2 3 4 0

x y z

x y z

x y z

− + =− + = −− + =


3 6227 3 2 1

3 042

−− −−

( )2 23 3

11 13

1 2

7 3 2 1

3 042

R r

−→ − =−

−

( )

2 23 3

2 1 25 83 3

3 1 35 83 3

2 23 3

5 83 2 33 3

1 27

0 15 2

0 4

1 2

0 15

0 0 0 19

R r r

R r r

R r r

−= − +−→ −= − +

− −

−

−→ − = +−




57. 6

3 2 5

3 2 14

x y z

x y z

x y z

+ − =− + = −+ − =


61 1 123 51

231 14

−− −

−

( )

2 1 2

3 1 3

234 15 5 2 25

715 5

234 1 2 15 5

3 2 36355

61 1 13

230 5 4

0 82 1

61 1 1

0 1

0 82 1

01

0 12

0 0

R r r

R r r

R r

R r r

R r r

−= − +

−→ −= − +

−−−→ = −−

−= − +−→= − +−

( )71

5 5

234 55 5 3 33

11 3 15

42 3 25

01

0 120 0 1

0 01 1

0 0 3 120 0 1

R r

R r r

R r r

−

−→ =−

= +→

= +−

The solution is 1, 3, 2x y z= = = − , or (1, 3, 2)− .

58. 4

2 3 4 15

5 2 12

x y z

x y z

x y z

− + = −− + = −+ − =


( )

2 1 2

3 1 3

2 2

1 1 1 43 152 4

25 1 12

1 1 1 42

0 7 1 25

0 6 7 32

1 1 1 40 7 1 20 6 7 32

R r r

R r r

R r

− −− −

−

− − = − +→ −−

= − +−

− −→ = −−

−

( )

1 2 1

3 2 3

13 35

1 3 1

2 3 2

0 31 1

0 7 1 26

0 0 5 10

0 31 1

0 7 1 20 0 1 2

0 01 1

0 0 3 12

0 0 1 2

R r r

R r r

R r

R r r

R r r

−= +

→ −= − +

−

−→ =−

−

= +→

= +−

The solution is 1, 3, 2x y z= = = − or (1, 3, 2)− .

59. 2 3

2 4 7

2 2 3 4

x y z

x y z

x y z

+ − = −− + = −

− + − =


31 2 14 72 1

2 32 4

−−− −

− −

( )

2 1 2

3 1 3

3 1 18 8 2 28

31 2 12

80 3 12

20 6 5

31 2 1

0 120 6 5

R r r

R r r

R r

−−= − +

−→ −= +

−−

−−−→ = −

−−

( )

1314 43 1 1 2 18 8

3 2 311 114 4

1314 43 1 48 8 3 311

11 3 11 4

2 32 3 28

012

0 16

0 0

01

0 1

0 0 1 1

0 0 31

0 0 1

0 0 1 1

R r r

R r r

R r

R r r

R r r

− −= − +−→= − +− −

− −

−→ = −

−= +

→= +

The solution is 1

3, , 12

x y z= − = = or 1

3, ,12

− .


1055


60. 4 3 8

3 3 12

6 1

x y z

x y z

x y z

+ − = −− + =+ + =


31 4 83 31 12

61 1 1

− −−

( )

2 1 2

3 1 3

3612 12 213 13 13

9 4013 13

1 2 1361213 13

3 2 381 913 13

31 4 83

0 13 36 12

0 3 9 9

31 4 8

0 1

0 3 9 9

014

0 13

0 0

R r r

R r r

R r

R r r

R r r

− − = − +→ −

= − +−

− −− −→ = −

−

= − +− −→= +

( )9 40

13 133612 13

3 313 13 8119

91 3 1138

3 123 3 2131

9

01

0 1

0 0 1

0 0 31

0 0 1

0 0 1

R r

R r r

R r r

−

− −→ =

= − +−→

= +

The solution is 8 1

3, ,3 9

x y z= = − = or 8 1

3, ,3 9

− .

61.

23

32 1

84 2

3

x y z

x y z

x y

+ − =

− + =

+ =


( )

23

83

1 1 23 3 9

11 13

83

3 1 1

2 1 1 1

04 2

1

2 1 1 1

04 2

R r

−−

−→ − =

( )

1 1 23 3 9

5 5 5 2 1 23 3 9

3 1 3162 43 3 9

1 1 23 3 9

1 33 2 25

162 43 3 9

13 1

1 2 11 33 2

3 2 33

12

0 4

0

1

0 1 1

0

0 01

0 1 1

0 0 2 2

R r r

R r r

R r

R r r

R r r

−= − +−→= − +

−

−→ = −−

= − +−→ −

= − +

( )

( )

13

1 13 3 32

13

23 2 3 2

0 01

0 1 1

0 0 1 1

0 01

0 0 1

0 0 1 1

R r

R r r

−→ =−

→ = +

The solution is 1 2

, , 13 3

x y z= = = or 1 2

, , 13 3

.

62. 83

1

2 1

2

x y

x y z

x y z

+ =− + =+ + =


83

1 1 0 1

2 1 1 1

1 2 1

−

( )

2 1 2

3 1 353

53

2 3

53 2 33

1 1 0 12

0 3 1 1

0 1 1

1 1 0 1Interchange

0 1 1 and

0 3 1 1

1 1 0 1

0 1 1 3

0 0 4 4

R r r

R r r

r r

R r r

= − +→ − −

= − +

→

− −

→ = +



( )

( )

( )

153 343

22 2 33

13

21 1 23

1 1 0 1

0 1 1

0 0 1 1

1 1 0 1

0 1 0

0 0 1 1

1 0 0

0 1 0

0 0 1 1

R r

R r r

R r r

→ =

→ = −

→ = −

The solution is 1 2

, , 13 3

x y z= = = or 1 2

, , 13 3

.

63.

4

2 0

3 2 6

2 2 2 1

x y z w

x y z

x y z w

x y z w

+ + + =− + =

+ + − =− − + = −


1 1 1 1 4

0 02 1 1

3 62 1 12 21 2 1

−−

− − −

2 1 2

3 1 3

4 1 4

1 1 1 1 4 22 80 3 1 3

2 4 60 1

0 3 3 51

R r r

R r r

R r r

= − +− −− −→ = − +

− − −− = − +− − −

( )

2 3

2 2

1 2 1

3 2 3

4 2 4

1 1 1 1 4Interchange2 4 60 1

and 2 80 3 1

0 3 3 51

1 1 1 1 4

0 61 2 4 2 80 3 1

0 3 3 51

20 31 1

0 61 2 4 30 0 5 10 10 30 0 3 13 13

r r

R r

R r r

R r r

R r r

− − −−→− −− −

− − −

→ = −− −− −

− − −−−− = − +

→ = += +

( )13 35

1 3 1

2 3 2

4 3 4

20 31 1

0 61 2 4 0 0 1 2 2

0 0 3 13 13

0 0 01 1

0 0 01 2 20 0 1 2 2 30 0 0 7 7

R r

R r r

R r r

R r r

−−−

→ =

− = +→ = − +

= − +

( )14 47

1 4 1

3 4 3

0 0 01 1

0 0 01 2 0 0 1 2 2

0 0 0 1 1

0 0 01 1

0 0 01 2 20 0 0 01

0 0 0 1 1

R r

R r r

R r r

−

→ =

= +→

= − +

The solution is 1, 2, 0, 1x y z w= = = = or (1, 2, 0, 1).

64.

4

2 0

2 3 6

2 2 2 1

x y z w

x y z

x y z w

x y z w

+ + + =− + + =

+ + − =− + − + = −


1 1 1 1 4

0 01 2 1

3 62 1 1

1 2 12 2

−−

−− −

2 1 2

3 1 3

4 1 4

2 3

1 2 1

3

1 1 1 1 4

0 3 2 1 4 20 31 1 2 20 3 0 74

1 1 1 1 4Interchange0 31 1 2

and 0 3 2 1 4

0 3 0 74

0 61 2 4

0 31 1 2 0 0 5 10 10

0 0 3 13 13

R r r

R r r

R r r

r r

R r r

R

= +→ = − +

−− − = +

−− −→

= − +−− −→ = − 2 3

4 2 4

3

3

r r

R r r

+= − +


1057


( )4 3 4

14 35

14 435

0 61 2 4

0 31 1 2 3 50 0 5 10 10

0 0 0 35 35

0 61 2 4

0 31 1 2 0 0 1 2 2

0 0 0 1 1

R r r

R r

R r

−− −→ = −

− −

=−− −→= −

Write the matrix as the corresponding system: 2 4 6

3 2

2 2

1

x z w

y z w

z w

w

+ + =− − = −

+ ==

Substitute and solve: 2(1) 2

2 2

0

z

z

z

+ =+ =

=

0 3(1) 2

3 2

1

y

y

y

− − = −− = −

=

2(0) 4(1) 6

0 4 6

2

x

x

x

+ + =+ + =

=

The solution is 2x = , 1y = , 0z = , 1w = or (2, 1, 0, 1).

65. 2 1

2 2 2

3 3 3

x y z

x y z

x y z

+ + =− + =+ + =


1 2 1 1

2 1 2 2

3 3 31

−

( )

2 1 2

3 1 3

3 2 3

1 2 1 12

0 5 0 0 3

0 5 0 0

1 2 1 1

0 5 0 0

0 0 0 0

R r r

R r r

R r r

= − +→ −

= − +−

→ − = − +

The matrix in the last step represents the system

2 1

5 0

0 0

x y z

y

+ + =− =

=

Substitute and solve: 5 0

0

y

y

− ==

2(0) 1

1

x z

z x

+ + == −

The solution is 0, 1 ,y z x= = − x is any real

number or {( , , ) | 0, 1 ,x y z y z x= = − x is any real number}.

66. 2 3

2 2 6

3 3 4

x y z

x y z

x y z

+ − =− + =− + =


31 2 162 1 2

3 31 4

−−−

( )

2 1 2

3 1 3

3 2 3

31 2 12

0 5 0 40 5 4 1

31 2 10 5 0 40 0 0 1

R r r

R r r

R r r

−= − +

→ −= − +

−

−→ − = − +


67. 5

3 2 2 0

x y z

x y z

− + =+ − =

Write the augmented matrix: 1 1 1 53 2 2 0

−−

( )

( )

( )

2 1 2

12 25

1 2 1

1 1 1 5 3

0 5 5 15

1 1 1 5

0 1 1 3

1 0 0 2

0 1 1 3

R r r

R r

R r r

−→ = − +

− −

−→ =

− −

→ = +− −

The matrix in the last step represents the system 2

3

x

y z

=− = −

or, equivalently, 2

3

x

y z

== −

Thus, the solution is 2x = , 3y z= − , z is any

real number or {( , , ) | 2, 3,x y z x y z= = − z is any real number}.



68. 2 4

3 1

x y z

x y z

+ − =− + + =

Write the augmented matrix: 2 1 1 4

1 1 3 1

−−

( )

( )

( )

1 2

2 1 2

12 25 3

3

43

1 2 153

interchange1 1 3 1

and 2 1 1 4

1 1 3 1 2

0 3 5 6

1 1 3 1

0 1 2

1 0 1

20 1

r r

R r r

R r

R r r

− − −→

−−

− − −→ = − +

− − −→ =

−→ = +


13 5

23

x z

y z

− =

+ =or, equivalently,

41

3 5

23

x z

y z

= +

= −

Thus, the solution is: 4

13

x z= + , 5

23

y z= − , z

is any real number or 4

( , , ) 1 ,3

x y z x z= +

52 , is any real number

3y z z= − .

69.

2 3 300

3 5

x y zx y zx y z

x y z

+ − =− − =

− + + =+ + =

Write the augmented matrix: 2 3 1 31 1 1 01 1 1 0

1 1 3 5

−− −

−

1 2

2 1 2

3 1 3

4 1 4

1 1 1 0interchange2 3 1 3

and 1 1 1 0

1 1 3 5

1 1 1 02

0 5 1 3

0 0 0 00 2 4 5

r r

R r rR r rR r r

− −−

→−

− − = − +→ = +

= − +

3 4

1 1 1 0

interchange0 5 1 3

and 0 2 4 5

0 0 0 0

r r

− −

→

( )

( )

2 3 2

1 2 1

3 2 3

13 319 18

18

1 1 1 00 1 7 7

20 2 4 50 0 0 0

1 0 8 70 1 7 7

20 0 18 19

0 0 0 0

71 0 870 1 7

=0 0 10 0 0 0

R r r

R r r

R r r

R r

− −− −

→ = − +

− −= +− −

→= − +

−−−−

→


7 7

1918

x z

y z

z

− = −− = −

=

Substitute and solve: 19

7 718

718

y

y

− =

=

198 7

18139

x

x

− = −

=

Thus, the solution is 139

x = , 7

18y = ,

1918

z = or

13 7 19, ,

9 18 18.

70.

3 1

2 4 0

3 2 1

2 5

x y z

x y z

x y z

x y

− + =− − =

− + =− =


2 1 4 0

1 3 2 1

1 2 0 5

−− −−−


1059


2 1 2

3 1 2

4 1 2

2 4

1 2 1

4 2

1 3 1 12

0 5 6 2

0 0 1 0

0 1 1 4

1 3 1 1

interchange0 1 1 4

and 0 0 1 0

0 5 6 2

1 0 2 13

30 1 1 4

0 0 1 0 5

0 0 1 22

R r r

R r r

R r r

r r

R r r

R r

−= +

− −→ = − +

= − +−

−−

→

− −

−= +−

→= −

− −4r+

1 3 1

2 3 2

4 3 4

1 0 0 132

0 1 0 4

0 0 1 0

0 0 0 22

R r r

R r r

R r r

= +→ = +

= +−


71. 4 4

2 3 3

x y z w

x y z w

+ + − =− + + =


1 2

4 1 1 1 4

1 1 2 3 3

interchange1 1 2 3 3

and 4 1 1 1 4 r r

−−

−→

−

( )2 1 21 1 2 3 3

40 5 7 13 8

R r r−

→ = − +− − −

The matrix in the last step represents the system 2 3 3

5 7 13 8

x y z w

y z w

− + + =− − = −

The second equation yields 5 7 13 8

5 7 13 8

7 13 85 5 5

y z w

y z w

y z w

− − = −= + −

= + −

The first equation yields 2 3 3

3 2 3

x y z w

x y z w

− + + == + − −

Substituting for y: 8 7 13

3 2 35 5 5

3 2 75 5 5

x z w z w

x z w

= + − + + − −

= − − +

Thus, the solution is 3 2 75 5 5

x z w= − − + ,

7 13 85 5 5

y z w= + − , z and w are any real numbers or

3 2 7 7 13 8( , , , ) , ,

5 5 5 5 5 5x y z w x z w y z w= − − + = + −

and are any real numbersz w .

72. 4 5

2 54

x yx y z w

z w

− + =− + − =

+ =

Write the augmented matrix: 4 1 0 0 5

2 1 1 1 50 0 1 1 4

−− −

( )

( )

( )

514 4

11 14

5144

1512 1 22 2

514 4

2 2

1 0 0

2 1 1 1 5 0 0 1 1 4

1 0 0

0 1 1 2

0 0 1 1 4

1 0 0

0 1 2 2 15 20 0 1 1 4

R r

R r r

R r

− −→ − − = −

−−→ − − = − +

− −→ − − = −

( )1 12 2

11 2 14

1 0 50 1 2 2 15 0 0 1 1 4

R r r

− −→ − − = +

11 3 12

2 3 2

1 0 0 1 30 1 0 4 7

20 0 1 1 4

R r r

R r r

− = +→ −

= +


4 7 4

x wy w

z w

+ = −+ = −

+ = or, equivalently,

37 4

4

x wy wz w

= − −= − −= −

The solution is 3x w= − − , 7 4y w= − − , 4z w= − ,

w is any real number or {( , , , ) | 3 ,x y z w x w= − −

7 4 , 4 , is any real number}y w z w w= − − = −



73. Each of the points must satisfy the equation 2y ax bx c= + + .

(1,2) : 2

( 2, 7) : 7 4 2

(2, 3) : 3 4 2

a b c

a b c

a b c

= + +− − − = − +

− − = + +

Set up a matrix and solve:

( )

2 1 2

3 1 3

1512 262 2

1 12 2

1 2 1512 2

3 2 3

1 1 1 22 74 1

34 2 1

1 1 1 2 40 3 15 6 4

20 3 11

1 1 1 2

0 120 3 11

01

0 1 22 60 0

R r r

R r r

R r

R r r

R r r

− −−

= − +→ − −− = − +

− − −

= −→− − −

−= − +

→ = +− −

( )1 12 2

11 53 322 2

11 3 12

12 3 22

1 0

0 1

0 0 1 3

1 0 0 2

0 1 0 1

0 0 1 3

R r

R r r

R r r

−→ → = −

− = − +→

= − +

The solution is 2, 1, 3a b c= − = = ; so the

equation is 22 3y x x= − + + .

74. Each of the points must satisfy the equation 2y ax bx c= + + .

(1, 1) : 1

(3, 1) : 1 9 3

(– 2,14) : 14 4 2

a b c

a b c

a b c

− − = + +− − = + +

= − +


2 1 2

3 1 3

1 1 1 1

9 3 1 1

4 1 142

1 1 1 1 90 8 6 8 40 3 186

R r r

R r r

−−

−

− = − +→ − − = − +

−−

12 264 4

3 33 2 3

1 13 3

1 2 14 43 3 1

3 35

11 3 13

41 3 23

1 1 1 1

0 1

0 0 5 10

01

0 1

0 0 1 2

0 01 1

0 0 1 40 0 1 2

R r

R r r

R r r

R r

R r r

R r r

−= −

−→= − +

−= − +

−→=

= +→ −= − +

The solution is 1, – 4, 2a b c= = = ; so the

equation is 2 4 2y x x= − + .

75. Each of the points must satisfy the equation 3 2( )f x ax bx cx d= + + + .

( 3) 112 : 27 9 3 112

( 1) 2 : 2

(1) 4 : 4

(2) 13 : 8 4 2 13

f a b c d

f a b c d

f a b c d

f a b c d

− = − − + − + = −− = − − + − + = −

= + + + == + + + =


27 9 3 1 11221 1 1 1

1 1 1 1 4

8 134 2 1

− − −−− −

3 1

1 1 1 1 4Interchange21 1 1 1

and 27 9 3 1 112

8 134 2 1

r r−− −→

− − −

2 1 2

3 1 3

4 1 4

1 1 1 1 4

0 02 2 2 2740 36 2824 8

4 60 7 19

R r r

R r r

R r r

= +→ = +

− = − +− − − −

( )12 22

1 2 1

3 2 3

4 2 4

1 1 1 1 4

0 01 1 1 40 36 2824

4 60 7 19

0 0 31 1

0 01 1 1 368 400 0 24 4

60 0 3 15

R r

R r r

R r r

R r r

→ =−

− − − −

= − +→ = − +

− − = +− − −


1061


( )13 3245 5

3 3

0 0 31 1

0 01 1 1

0 0 160 0 3 15

R r→ =− −

− − −

( )

1 143 3

1 3 1

51 4 3 43 3

1 143 3

14 4551

3 3

11 4 13

2 4 2

13 4 33

0 01

0 01 1 1 60 0 1

250 0 0 5

0 01

0 01 1 1 0 0 1

0 0 0 51

0 0 0 3140 0 01

0 0 0 01

0 0 0 51

R r r

R r r

R r

R r r

R r r

R r r

= − +→

= +− −−−

→ = −− −

= − +−→ = − +

= +

The solution is 3, 4, 0, 5a b c d= = − = = ; so the

equation is 3 2( ) 3 4 5f x x x= − + .

76. Each of the points must satisfy the equation 3 2( )f x ax bx cx d= + + + .

( 2) 10 : 8 4 2 10

( 1) 3 : 3

(1) 5 : 5

(3) 15 : 27 9 3 15

f a b c d

f a b c d

f a b c d

f a b c d

− = − − + − + = −− = − + − + =

= + + + == + + + =


3 1

2 1 2

3 1 3

4 1 4

104 18 231 1 1 1

51 1 1 1

27 9 3 151

51 1 1 1Interchange31 1 1 1

and 104 18 227 9 3 151

51 1 1 1

0 0 82 2 80 6 9 3012 270 18 12024 26

r r

R r r

R r r

R r r

−− −− −

− −→−− −

= +→ = +

= − +− −− −

( )12 22

51 1 1 1

0 01 1 4 0 6 9 3012

0 18 12024 26

R r→ =

− −− −

( )

1 2 1

3 2 3

4 2 4

13 361

2

0 01 1 1

0 01 1 4 120 0 6 3 18 180 0 24 8 48

0 01 1 1

0 01 1 4

0 0 31

0 0 24 8 48

R r r

R r r

R r r

R r

= − +→ = − +

− − = +− − −

→ =− −

− − −

( )

12

1 3 1

1 4 3 42

12

14 4201

2

11 4 12

2 4 21

3 4 32

0 01 4

0 01 1 4 24

0 0 31

0 0 0 12020

0 01 4

0 01 1 4 0 0 31

0 0 0 61

0 0 01 1

0 0 01 2 0 0 0 01

0 0 0 61

R r r

R r r

R r

R r r

R r r

R r r

= − +→

= +− −−−

→ = −− −

= − +−→ = − +

= +

The solution is 1, 2, 0, 6a b c d= = − = = ; so the

equation is 3 2( ) 2 6f x x x= − + .

77. Let x = the number of servings of salmon steak. Let y = the number of servings of baked eggs. Let z = the number of servings of acorn squash. Protein equation: 30 15 3 78x y z+ + = Carbohydrate equation: 20 2 25 59x y z+ + = Vitamin A equation: 2 20 32 75x y z+ + = Set up a matrix and solve:

( )

3 1

11 12

30 15 3 78

20 25 592

20 32 752

20 32 752Interchange

20 25 59 2 and r

30 15 3 78

10 16 37.51

20 25 59 2

30 15 3 78

r

R r

→

→ =

2 1 2

3 1 3

10 16 37.5120

0 198 295 691 30

0 285 477 1047

R r r

R r r

= − +→ − − −

= − +− − −



( )

( )

953 2 366

3457 345766 66

663 33457

10 16 37.51

0 198 295 691

0 0

10 16 37.51

0 198 295 691

0 0 1 1

R r r

R r

→ − − − = − +− −

→ − − − = −

Substitute 1z = and solve: 198 295(1) 691

198 396

2

y

y

y

− − = −− = −

=

10(2) 16(1) 37.5

36 37.5

1.5

x

x

x

+ + =+ =

=

The dietitian should serve 1.5 servings of salmon steak, 2 servings of baked eggs, and 1 serving of acorn squash.

78. Let x = the number of servings of pork chops. Let y = the number of servings of corn on the cob. Let z = the number of servings of 2% milk. Protein equation: 23 3 9 47x y z+ + = Carbohydrate equation: 16 13 58y z+ = Calcium equation: 10 10 300 630x y z+ + = Set up a matrix and solve:

13 110

3 1 313 2916 8 1

2 216

467 47516 813 2916

23 3 9 47

0 16 13 58

10 10 300 630

30 631 1 Interchange0 16 13 58

and 23 3 9 47

30 631 123

0 1

0 140220 681

1 0

0 1

r r

R r r

R r

→

= − +→

=−− −

→

( )

1 2 18

3 2 32659

4

467 47516 813 29 4

3 316 8 2659

4671 3 116

132 3 216

20

0 0 1402

1 0

0 1

0 0 1 2

1 0 0 1

0 1 0 2

0 0 1 2

R r r

R r r

R r

R r r

R r r

= − += +

− −

→ = −

= − +→

= − +

The dietitian should provide 1 serving of pork chops, 2 servings of corn on the cob, and 2 servings of 2% milk.

79. Let x = the amount invested in Treasury bills. Let y = the amount invested in Treasury bonds. Let z = the amount invested in corporate bonds. Total investment equation: 10,000x y z+ + = Annual income equation: 0.06 0.07 0.08 680x y z+ + = Condition on investment equation: 0.5

2 0

z x

x z

=− =

Set up a matrix and solve: 10,0001 1 1

0.06 0.07 0.08 680

0 021 −

( )

2 1 2

3 1 3

2 2

10,0001 1 10.06

0 0.01 0.02 80 10,0000 31

10,0001 1 1

0 8000 1001 210,0000 31

R r r

R r r

R r

= − +→

= − +−−−

→ =−−−

( )

1 2 1

3 2 3

3 3

1 3 1

2 3 2

0 20001 1

0 8000 1 2

0 0 20001

0 20001 1

0 8000 1 2

0 0 20001

0 0 40001

0 0 4000 12

0 0 20001

R r r

R r r

R r

R r r

R r r

−= − +

→= +

−−

−→ = −

= +→

= − +

Carletta should invest $4000 in Treasury bills, $4000 in Treasury bonds, and $2000 in corporate bonds.

80. Let x = the fixed delivery charge; let y = the cost of each tree, and let z = the hourly labor charge. 1st subdivision: 250 166 7520x y z+ + = 2nd subdivision: 200 124 5945x y z+ + = 3rd subdivision: 300 200 8985x y z+ + = Set up a matrix and solve:

250 166 75201

200 59451 124

300 200 89851


1063


2 1 2

3 3 1

12 250

13 350

250 166 75201

0 50 1575 42

0 50 34 1465

250 166 75201

0 0.84 31.5 1

0 0.68 29.31

R r r

R r r

R r

R r

= −→

= −

=→

=

( )

1 1 2

3 2 3

13 30.16

1 1 3

2 2 3

0 3551 44250

0 0.84 31.5 1

0 0 0.18 2.2

0 3551 44

0 0.84 31.5 1

0 0 13.751

0 0 250144

0 0 19.95 10.84

0 0 13.751

R r r

R r r

R r

R r r

R r r

−−= −

→= −

−−→ =

= +→

= −

The delivery charge is $250 per job, the cost for each tree is $19.95, and the hourly labor charge is $13.75.

81. Let x = the number of Deltas produced. Let y = the number of Betas produced. Let z = the number of Sigmas produced. Painting equation: 10 16 8 240x y z+ + = Drying equation: 3 5 2 69x y z+ + = Polishing equation: 2 3 41x y z+ + = Set up a matrix and solve:

( )

( )

1 2 1

2 1 2

3 1 3

12 22

10 16 8 240

3 5 2 69

2 3 1 41

1 1 2 33

3 5 2 69 3

2 3 1 41

1 1 2 333

0 2 4 30 2

0 1 3 25

1 1 2 33

0 1 2 15

0 1 3 25

1 0 4 48

0 1 2 15

0 0 1 10

R r r

R r r

R r r

R r

→ = − +

= − +→ − −

= − +− −

→ − − =− −

→ − −− −

1 1 2

3 3 2

R r r

R r r

= −= −

( )3 3

1 3 1

2 3 2

1 0 4 48

0 1 2 15

0 0 1 10

1 0 0 84

0 1 0 5 2

0 0 1 10

R r

R r r

R r r

→ − − = −

= − +→

= +

The company should produce 8 Deltas, 5 Betas, and 10 Sigmas.

82. Let x = the number of cases of orange juice produced; let y = the number of cases of grapefruit juice produced; and let z = the number of cases of tomato juice produced. Sterilizing equation: 9 10 12 398x y z+ + = Filling equation: 6 4 4 164x y z+ + = Labeling equation: 2 58x y z+ + = Set up a matrix and solve:

( )

1 3

2 1 2

3 1 3

1 12 24 8

9 10 39812

6 1644 4

581 2 1

581 2 1Interchange

6 164 4 4 and

9 10 39812

581 2 16

0 184 8 29

0 3 1248

581 2 1

0 23 1

0 3 1248

r r

R r r

R r r

R r

→

= − +→ −− −

= − +−−

→ = −−−

( )

12

1 2 114

3 2 3

121 1

3 34 5

01 122

0 23 18

0 0 5 60

01 12

0 23 1

0 0 1 12

R r r

R r r

R r

= − +→

= +

→ =

11 3 12

12 3 24

0 0 61

0 0 20 1

0 0 1 12

R r r

R r r

= − +→

= − +

The company should prepare 6 cases of orange juice, 20 cases of grapefruit juice, and 12 cases of tomato juice.



83. Rewrite the system to set up the matrix and solve:

2 2

4 1 1 4

3 1 1 3

3 4 1 1 3 4

4 8 2 0 2 4

8 5 5 8

4 3 3 4

0

I I

I I I I

I I I I

I I I I I I

− + − = == + + =

→= + + =

+ = − − =

2 1

12 22

3 1 3

4 1 4

0 0 02 4

0 0 5 81

0 3 01 4

0 01 1 1


and 0 3 01 4

0 01 1 1

0 0 5 81

0 0 01 2 40 0 3 5

6 80 0 1

r r

R r

R r r

R r r

− −

→

− −

=→ = − +

−− = − +− −−

( )

3 4

3 3

4 3 4

14 423

2823


and 6 80 0 140 0 3 5

0 0 5 81

0 0 01 2 30 0 6 81

23 280 0 0

0 0 5 81

0 0 01 2

0 0 6 81

0 0 0 1

r r

R r

R r r

R r

→− −−

−−

= −→

= − +− −

→ = −

4423

1 4 116

3 4 323

2823

0 0 0150 0 01 2

60 0 01

0 0 0 1

R r r

R r r

= − +→

= − +

The solution is 14423

I = , 2 2I = , 31623

I = ,

42823

I = .

84. Rewrite the system to set up the matrix and solve:

1 3 2 1 2 3

1 3 1 3

1 2 1 2

0

24 6 3 0 6 3 24

12 24 6 6 0 6 6 36

I I I I I I

I I I I

I I I I

= + − − =− − = → − − = −

+ − − = − − = −

( )

2 1 2

3 1 3

3 12 22 6

12

1 2 132

3 2 3

12

1 1 1 0

6 0 3 24

6 6 0 36

1 1 1 06

0 6 9 24 6

0 12 6 36

1 1 1 0

0 1 4

0 12 6 36

1 0 4

0 1 4 12

0 0 12 12

1 0 4

0 1

R r r

R r r

R r

R r r

R r r

− −− − −− − −

− −= +

→ − − −= +

− − −− −

→ = −− − −

= +→

= +

→ ( )3 13 32 12

11 3 12

32 3 22

4

0 0 1 1

1 0 0 3.5

0 1 0 2.5

0 0 1 1

R r

R r r

R r r

=

= − +→

= − +

The solution is 1 2 33.5, 2.5, 1I I I= = = .

85. Let x = the amount invested in Treasury bills. Let y = the amount invested in corporate bonds. Let z = the amount invested in junk bonds.

a. Total investment equation: 20,000x y z+ + = Annual income equation: 0.07 0.09 0.11 2000x y z+ + = Set up a matrix and solve:

( )

( )

( )

2 2

2 2 1

12 22

1 1 1 20,000

0.07 0.09 0.11 2000

1 1 1 20,000 100

7 9 11 200,000

1 1 1 20,000 7

0 2 4 60,000

1 1 1 20,000

0 1 2 30,000

R r

R r r

R r

→ =

→ = −

→ =


1065


( )1 1 21 0 1 10,000

0 1 2 30,000

R r r− −

→ = −

The matrix in the last step represents the

system 10,000

2 30,000

x z

y z

− = −+ =

Therefore the solution is 10,000x z= − + , 30,000 2y z= − , z is any real number.

Possible investment strategies:

Amount Invested At

7% 9% 11%

0 10,000 10,000

1000 8000 11,000

2000 6000 12,000

3000 4000 13,000

4000 2000 14,000

5000 0 15,000

b. Total investment equation:

25,000x y z+ + = Annual income equation: 0.07 0.09 0.11 2000x y z+ + = Set up a matrix and solve:

( )

( )

( )

( )

2 2

2 2 1

12 22

1 1 2

1 1 1 25,000

0.07 0.09 0.11 2000

1 1 1 25,000 100

7 9 11 200,000

1 1 1 25,000 7

0 2 4 25,000

1 1 1 25,000

0 1 2 12,500

1 0 112,500

0 1 2 12,500

R r

R r r

R r

R r r

→ =

→ = −

→ =

−→ = −


system 12,500

2 12,500

x z

y z

− =+ =

Thus, the solution is 12,500x z= + , 2 12,500y z= − + , z is any real number.


Amount Invested At

7% 9% 11%

12,500 12,500 0

14,500 8500 2000

16,500 4500 4000

18,750 0 6250

c. Total investment equation:

30,000x y z+ + = Annual income equation: 0.07 0.09 0.11 2000x y z+ + = Set up a matrix and solve:

( )

( )

( )

( )

2 2

1 2 1

12 22

1 1 2

1 1 1 30,000

0.07 0.09 0.11 2000

1 1 1 30,000 100

7 9 11 200,000

1 1 1 30,000 7

0 2 4 10,000

1 1 1 30,000

0 1 2 5000

1 0 1 35,000

0 1 2 5000

R r

R r r

R r

R r r

→ =

→ = −−

→ =−

−→ = −

−


system 35,000

2 5000

x z

y z

− =+ = −

Thus, the solution is 35,000x z= + , 2 5000y z= − − , z is any real number.

However, y and z cannot be negative. From 2 5000y z= − − , we must have 0.y z= =

One possible investment strategy

Amount Invested At

7% 9% 11%

30,000 0 0

This will yield ($30,000)(0.07) = $2100,

which is more than the required income.

d. Answers will vary.



86. Let x = the amount invested in Treasury bills. Let y = the amount invested in corporate bonds. Let z = the amount invested in junk bonds. Let I = income Total investment equation: 25,000x y z+ + = Annual income equation: 0.07 0.09 0.11x y z I+ + = Set up a matrix and solve:

( )

( )

( )

( )

2 2

1 2 1

12 22

1 1 2

1 1 1 25,000

0.07 0.09 0.11

1 1 1 25,000 100

7 9 11 100

1 1 1 25,000 7

0 2 4 100 175,000

1 1 1 25,000

0 1 2 50 87,500

1 0 1112,500 50

0 1 2 50 87,500

I

R rI

R r rI

R rI

IR r r

I

→ =

→ = −−

→ =−

− −→ = −

−

The matrix in the last step represents the system 112,500 50

2 50 87,500

x z I

y z I

− = −+ = −

Thus, the solution is 112,500 50x I z= − + , 50 87,500 2y I z= − − , z is any real number.

a. 1500I =

( )112,500 50

112,500 50 1500

37,500

x I z

z

z

= − += − += +

( )50 87,500 2

50 1500 87,500 2

12,500 2

y I z

z

z

= − −= − −= − −

z is any real number. Since y and z cannot be negative, we must have 0.y z= = Investing all of the money at 7% yields $1750, which is more than the $1500 needed.

b. 2000I =

( )112,500 50

112,500 50 2000

12,500

x T z

z

z

= − += − += +

( )50 87,500 2

50 2000 87,500 2

12,500 2

y I z

z

z

= − −= − −= −

z is any real number.


Amount Invested At

7% 9% 11%

12,500 12,500 0 15,500 6500 3000 18,750 0 6250

c. 2500I =

( )112,500 50

112,500 50 2500

12,500

x T z

z

z

= − += − += − +

( )50 87,500 2

50 2500 87,500 2

37,500 2

y I z

z

z

= − −= − −= −

z is any real number.


Amount invested at

7% 9% 11%

0 12,500 12,500 1000 10,500 13,500 6250 0 18,750

d. Answers will vary. 87. Let x = the amount of supplement 1.

Let y = the amount of supplement 2. Let z = the amount of supplement 3.

0.20 0.40 0.30 40 Vitamin C

0.30 0.20 0.50 30 Vitamin D

x y z

x y z

+ + =+ + =

Multiplying each equation by 10 yields 2 4 3 400

3 2 5 300

x y z

x y z

+ + =+ + =


( )

( )

( )

312

1 12

32

2 2 112

312

2 21 48

2 4 3 400

3 2 5 300

1 2 200

3 2 5 300

1 2 200 3

0 4 300

1 2 200

0 1 75

R r

R r r

R r

→ =

→ = −− −

→ = −−

Section 11.3: Systems of Linear Equations: Determinants

1067


( )74

1 1 218

1 0 50 2

0 1 75R r r→ = −

−

The matrix in the last step represents the system

74

18

50

75

x z

y z

+ =

− =

Therefore the solution is 7

504

x z= − ,

175

8y z= + , z is any real number.

Possible combinations:

Supplement 1 Supplement 2 Supplement 3

50mg 75mg 0mg

36mg 76mg 8mg

22mg 77mg 16mg

8mg 78mg 24mg

88. Let x = the amount of powder 1.

Let y = the amount of powder 2. Let z = the amount of powder 3.

120.20 0.40 0.30 12 Vitamin B

0.30 0.20 0.40 12 Vitamin E

x y z

x y z

+ + =+ + =

Multiplying each equation by 10 yields 2 4 3 120

3 2 4 120

x y z

x y z

+ + =+ + =


( )

( )

32 2 12

1 1 2

2 4 3 120

3 2 4 120

2 4 3 120

0 4 0.5 60

2 0 2.5 60

0 4 0.5 60

R r r

R r r

→ = −− − −

→ = +− − −

The matrix in the last step represents the system 2 2.5 60

4 0.5 60

x z

y z

+ =− − = −

Thus, the solution is 30 1.25x z= − , 15 0.125y z= − , z is any real number.

Possible combinations:

Powder 1 Powder 2 Powder 3

30 units 15 units 0 units




89 – 91. Answers will vary. Section 11.3

1. ad bc−

2. 5 3

3 4− −

3. False; If ad=bc, then the det = 0.

4. False; The solution cannot be determined.

5. False; See Thm (15)

6. False; See Thm (14)

7. 6 4 6(3) ( 1)(4) 18 4 2231

= − − = + =−

8. 8 3 8(2) 4( 3) 16 12 284 2

− = − − = + =

9. 3 1 3(2) 4( 1) 6 4 24 2

− − = − − − = − + = −

10. 24 4(3) ( 5)(2) 12 10 25 3

− = − − − = − + = −−

11.

( ) [ ][ ]

3 4 25 51 1 1 151 1 3 4 22 22 1 1 221 2

3 1 ( 2) 2(5) 4 1( 2) 1(5)

2 1(2) 1( 1)

3( 8) 4( 7) 2(3)

24 28 6

10

− −− = − +− −−

= − − − − − −

+ − −= − − − += − + +=


1068


12.

[ ] [ ][ ]

31 25 6 5 61 16 5 1 3 ( 2)13 8 3 82 28 32

1 1(3) 2( 5) 3 6(3) 8( 5)

2 6(2) 8(1)

1(13) 3(58) 2(4)

13 174 8

169

−− −− = − + −

= − − − − −− −

= − −= − −= −

13.

[ ] [ ][ ]

4 1 20 6 0 61 16 0 4 ( 1) 21

3 34 1 4 131 4

4 1(4) 0( 3) 1 6(4) 1(0)

2 6( 3) 1( 1)

4( 4) 1(24) 2( 17)

16 24 34

26

−− −= − − +−− −−

= − − − + −+ − − −

= − + + −= − + −= −

14.

[ ] [ ][ ]

3 9 40 04 1 1 401 4 3 ( 9) 4

3 8 8 31 18 3 1

3 4(1) ( 3)(0) 9 1(1) 8(0)

4 1( 3) 8(4)

3(4) 9(1) 4( 35)

12 9 140

119

−= − − +

− −−

= − − + −+ − −

= + + −= + −= −

15. 8

4

x y

x y

+ =− =

1 1 1 1 21 1

8 1 8 4 124 1

81 4 8 41 4

x

y

D

D

D

= = − − = −−

= = − − = −−

= = − = −

Find the solutions by Cramer's Rule: 12 4

6 22 2

yxDD

x yD D

− −= = = = = =− −

The solution is (6, 2).

16. 2 5

3

x y

x y

+ =− =

1 2 1 2 31 1

5 2 5 6 113 1

51 3 5 231

x

y

D

D

D

= = − − = −−

= = − − = −−

= = − = −

Find the solutions by Cramer's Rule: 11 11 2 23 3 3 3

yxDD

x yD D

− −= = = = = =− −

The solution is 11 2,

3 3.

17. 5 13

2 3 12

x y

x y

− =+ =

5 1 15 2 1732

13 1 39 12 51312

5 13 60 26 342 12

x

y

D

D

D

−= = + =

−= = + =

= = − =


3 217 17

yxDD

x yD D

= = = = = =


18. 3 5

2 3 8

x y

x y

+ =− = −

31 3 6 932

5 3 15 ( 24) 98 3

51 8 10 1882

x

y

D

D

D

= = − − = −−

= = − − − =− −

= = − − = −−


1 29 9

yxDD

x yD D

−= = = − = = =− −

The solution is ( 1, 2)− .


1069


19. 3 24

2 0

x

x y

=+ =

3 0 6 0 61 2

024 48 0 480 2

3 24 0 24 2401

x

y

D

D

D

= = − =

= = − =

= = − = −


8 46 6

yxDD

x yD D

−= = = = = = −

The solution is (8, 4)− .

20. 4 5 3

2 4

x y

y

+ = −− = −

54 8 0 80 2

3 5 6 ( 20) 264 2

34 16 0 160 4

x

y

D

D

D

= = − − = −−

−= = − − =− −

−= = − − = −−

Find the solutions by Cramer's Rule: 26 13 16

28 4 8

yxDD

x yD D

−= = = − = = =− −

The solution is 13

, 24

− .

21. 3 6 24

5 4 12

x y

x y

− =+ =

3 6 12 ( 30) 425 4

624 96 ( 72) 16812 4

3 24 36 120 845 12

x

y

D

D

D

−= = − − =

−= = − − =

= = − = −


4 242 42

yxDD

x yD D

−= = = = = = −


22. 2 4 16

3 5 9

x y

x y

+ =− = −

2 4 10 12 223 5

16 4 80 36 449 5

x

D

D

= = − − = −−

= = − + = −− −

162 18 48 663 9

yD = = − − = −−


2 322 22

yxDD

x yD D

− −= = = = = =− −


23. 3 2 4

6 4 0

x y

x y

− =− =

23 12 ( 12) 046

D −= = − − − =−

Since 0D = , Cramer's Rule does not apply.

24. 2 5

4 8 6

x y

x y

− + =− =

1 2 8 8 04 8

D −= = − =−


25. 2 4 2

3 2 3

x y

x y

− = −+ =

42 4 12 163 2

2 4 4 12 83 2

22 6 6 123 3

x

y

D

D

D

−= = + =

− −= = − + =

−= = + =

Find the solutions by Cramer's Rule: 8 1 12 3

16 2 16 4yx

DDx y

D D= = = = = =

The solution is 1 3

,2 4

.


1070


26. 3 3 3

84 2

3

x y

x y

+ =

+ =

83

83

3 3 6 12 64 2

3 36 8 2

2

3 38 12 4

4

x

y

D

D

D

= = − = −

= = − = −

= = − = −


yxDD

x yD D

− −= = = = = =− −

The solution is 1 2

,3 3

.

27. 2 3 1

10 10 5

x y

x y

− = −+ =

32 20 ( 30) 5010 10

31 10 ( 15) 55 10

2 1 10 ( 10) 2010 5

x

y

D

D

D

−= = − − =

−−= = − − − =

−= = − − =

Find the solutions by Cramer's Rule: 5 1 20 2

50 10 50 5yx

DDx y

D D= = = = = =

The solution is 1 2

,10 5

.

28. 3 2 0

5 10 4

x y

x y

− =+ =

3 2 30 ( 10) 405 10

0 2 0 ( 8) 8104

3 0 12 0 125 4

x

y

D

D

D

−= = − − =

−= = − − =

= = − =


yxDD

x yD D

= = = = = =

The solution is 1 3

,5 10

.

29. 2 3 6

12

x y

x y

+ =

− =

12

12

32 2 3 51 1

6 3 3 156

2 21

621 6 5

1

x

y

D

D

D

= = − − = −−

= = − − = −−

= = − = −

Find the solutions by Cramer's Rule: 15

3 52 15 2 5

yx DDx y

D D

− −= = = = = =− −

The solution is 3,

21 .

30. 1

22

2 8

x y

x y

+ = −

− =

12

12

1 1 1 21 2

2 1 4 8 48 2

2 4 ( 2) 681

x

y

D

D

D

= = − − = −−

−= = − = −−

−= = − − =


2 32 2

yxDD

x yD D

−= = = = = = −− −


31. 3 5 3

15 5 21

x y

x y

− =+ =

3 5 15 ( 75) 9015 5

3 5 15 ( 105) 120521

3 3 63 45 1815 21

x

y

D

D

D

−= = − − =

−= = − − =

= = − =


yxDD

x yD D

= = = = = =

The solution is 4 1,

3 5.


1071


32. 2 1

1 32 2

x y

x y

− = −

+ =

12

3 12 2

32

2 11 1 2

1

1 1 1 31

2 2

2 13 1 4

1

x

y

D

D

D

−= = + =

− −= = − + =

−= = + =


22 2

yxDD

x yD D

= = = = =

The solution is 1,

22 .

33. 6

3 2 5

3 2 14

x y z

x y z

x y z

+ − =− + = −+ − =

1 1 123 1

231

2 23 31 11 1 ( 1)2 23 31 1

1(4 3) 1( 6 1) 1(9 2)1 7 11

3

D−

−=−

− −= − + −− −

= − − − − − += + −= −

6 1 125 1

2314

2 25 51 16 1 ( 1)2 23 314 14

6(4 3) 1(10 14) 1( 15 28)6 4 13

3

xD−

−= −−

− −− −= − + −− −

= − − − − − += + −= −

61 1

3 5 121 14

5 3 3 51 11 6 ( 1)2 214 1 1 14

1(10 14) 6( 6 1) 1(42 5)

4 42 47

9

yD

−= −

−

− −= − + −− −

= − − − − − += − + −= −

61 123 5

31 14

2 25 3 5 31 1 63 314 1 14 1

1( 28 15) 1(42 5) 6(9 2)

13 47 66

6

zD −= −

− −− −= − +

= − + − + + += − − +=

Find the solutions by Cramer's Rule:

3 91 3

3 36

23

yx

z

DDx y

D DD

zD

− −= = = = = =− −

= = = −−

The solution is (1, 3, 2)− .

34. 4

2 3 4 15

5 2 12

x y z

x y z

x y z

− + = −− + = −+ − =

1 1 132 4

5 1 2

3 34 2 4 21 ( 1) 12 25 51 1

1(6 4) 1( 4 20) 1(2 15)2 24 17

5

D−−=

−

− −= − − +− −

= − + − − + += − += −

1 1415 3 412 1 2

3 15 15 34 44 ( 1) 12 21 12 12 1

4(6 4) 1(30 48) 1( 15 36)8 18 215

xD−−

= − −−

− − − −= − − − +− −

= − − + − + − += − − += −

1 14152 4

5 12 2

15 154 2 4 21 ( 4) 12 25 512 12

1(30 48) 4( 4 20) 1(24 75)

18 96 99

15

yD−

−=−

− −= − − +− −

= − + − − + += − − += −


1072


1 1 43 152

5 1 12

3 15 15 32 21 ( 1) ( 4)5 51 12 12 1

1( 36 15) 1(24 75) 4(2 15)21 99 68

10

zD− −− −=

− − − −= − − + −

= − + + + − += − + −=


5 151 3

5 510

25

yx

z

DDx y

D DD

zD

− −= = = = = =− −

= = = −−

The solution is (1, 3, 2)− .

35. 2 3

2 4 7

2 2 3 4

x y z

x y z

x y z

+ − = −− + = −

− + − =

1 2 142 1

2 32

4 41 2 1 21 2 ( 1)2 23 32 2

1(12 2) 2( 6 2) 1(4 8)10 8 422

3 2 147 1

34 2

4 47 71 13 2 ( 1)3 32 4 4 2

3(12 2) 2(21 4) 1( 14 16)30 34 266

x

D

D

−−=

− −

− −= − + −− −− −

= − − − + − −= + +=

− −−= −

−

− −− −= − − + −− −

= − − − − − − += − − −= −

31 1

72 1

2 34

7 71 2 1 21 ( 3) ( 1)2 23 34 4

1(21 4) 3( 6 2) 1(8 14)

17 12 6

11

yD

− −−=

− −

− −= − − + −− −− −

= − + − + − −= − +=

31 24 72

2 2 4

4 47 72 21 2 ( 3)2 22 4 4 2

1( 16 14) 2(8 14) 3(4 8)2 12 12

22

zD−

− −=−

− −− −= − + −− −

= − + − − − −= − + +=


3 22 22 2

221

22

yx

z

DDx y

D DD

zD

−= = = − = = =

= = =

The solution is 1

3, , 12

− .

36. 4 3 8

3 3 126 1

x y zx y z

x y z

+ − = −− + =

+ + =

31 43 31

61 1

3 3 3 31 11 4 ( 3)6 61 1 1 1

1( 6 3) 4(18 3) 3(3 1)9 60 1281

D−

= −

− −= − + −

= − − − − − += − − −= −

348312 161 1

3 31 12 12 18 4 ( 3)6 61 1 1 1

8( 6 3) 4(72 3) 3(12 1)72 276 39

243

xD−−

= −

− −= − − + −

= − − − − − − += − −= −

31 83 312

61 1

3 3 3 312 121 ( 8) ( 3)6 61 1 1 1

1(72 3) 8(18 3) 3(3 12)69 120 27216

yD−−

=

= − − + −

= − + − − −= + +=


1073


1 4 83 1 12

1 1 1

3 31 12 12 11 4 ( 8)1 1 1 1 1 1

1( 1 12) 4(3 12) 8(3 1)

13 36 32

9

zD−

= −

− −= − + −

= − − − − − += − + −= −


3 81 81 3

9 181 9

yx

z

DDx y

D D

Dz

D

−= = = = = = −− −−= = =−

The solution is 8 1

3, ,3 9

− .

37. 2 3 1

3 2 0

2 4 6 2

x y z

x y z

x y z

− + =+ − =− + =

2 3123 1

4 62

2 23 31 11 ( 2) 36 64 2 2 4

1(6 8) 2(18 4) 3( 12 2)

2 44 42

0

D

−−=

−

− −= − − +− −

= − + + + − −= − + −=


38. 2 5

3 2 4

2 2 4 10

x y z

x y

x y z

− + =+ =

− + − = −

1 1 2

3 02

22 4

0 3 0 32 21 ( 1) 22 24 2 4 2

1( 8 0) 1( 12 0) 2(6 4)

8 12 20

0

D

−=

− −

= − − +− − − −

= − − + − − + += − − +=


39. 2 0

2 4 0

2 2 3 0

x y z

x y z

x y z

+ − =− + =

− + − =

1 2 142 1

2 32

4 41 2 1 21 2 ( 1)2 23 32 2

1(12 2) 2( 6 2) 1(4 8)10 8 422

D−

−=− −

− −= − + −− −− −

= − − − + − −= + +=

0 2 140 0 [By Theorem (12)]1

0 32xD

−−= =

−

01 1

02 1 0 [By Theorem (12)]2 0 3

yD

−= =

− −

01 24 02 0 [By Theorem (12)]

2 02zD −= =

−


0 022 220

022

yx

z

DDx y

D DD

zD

= = = = = =

= = =

The solution is (0, 0, 0).

40. 4 3 0

3 3 0

6 0

x y z

x y z

x y z

+ − =− + =

+ + =

31 4

3 31

61 1

3 3 3 31 11 4 ( 3)6 61 1 1 1

1( 6 3) 4(18 3) 3(3 1)

9 60 12

81

0 34

0 3 0 [By Theorem (12)]1

0 61x

D

D

−= −

− −= − + −

= − − − − − += − − −= −

−= =−


1074


0 31

3 0 3 0 [By Theorem (12)]

0 61

01 4

3 0 0 [By Theorem (12)]1

01 1

y

z

D

D

−= =

= =−


0 00 0

81 81

00

81

yx

z

DDx y

D D

Dz

D

= = = = = =− −

= = =−

The solution is (0, 0, 0).

41. 2 3 0

3 2 0

2 4 6 0

x y z

x y z

x y z

− + =+ − =− + =

2 3123 1

4 62

2 23 31 11 ( 2) 34 46 62 2

1(6 8) 2(18 4) 3( 12 2)

2 44 42

0

D

−−=

−

− −= − − +− −

= − + + + − −= − + −=


42. 2 0

3 2 0

2 2 4 0

x y z

x y

x y z

− + =+ =

− + − =

1 1 2

3 02

22 4

0 3 0 32 21 ( 1) 22 24 2 4 2

1( 8 0) 1( 12 0) 2(6 4)

8 12 20

0

D

−=

− −

= − − +− − − −

= − − + − − + += − − +=


43. 4

1 2 3

x y z

u v w =

By Theorem (11), the value of a determinant changes sign if any two rows are interchanged.

Thus,

1 2 3

4u v w

x y z

= − .

44. 4

1 2 3

x y z

u v w =

By Theorem (14), if any row of a determinant is multiplied by a nonzero number k, the value of the determinant is also changed by a factor of k.

Thus, 2 2(4) 8

2 4 6 1 2 3

x y z x y z

u v w u v w= = = .

45. Let 4

1 2 3

x y z

u v w = .

3 6 9 3 1 2 3 [Theorem (14)]

3( 1) [Theorem (11)]

1 2 3

3(4) 12

x y z x y z

u v w u v w

x y z

u v w

− − − = −

= − −

= =

46. Let 4

1 2 3

x y z

u v w =

( )2 2 3

1 2 3 1 2 3[Theorem (15)]

( 1) 1 2 3 [Theorem (11)]

( 1)( 1) [Theorem (11)]

1 2 3

1 2 3

x u y v z w x y zR r r

u v w u v w

x y z

u v w

x y z

u v w

x y z

u v w

− − − == +

= −

= − −

= 4=


1075


47. Let 4

1 2 3

x y z

u v w =

1 2 3

3 6 9

2 2 2

1 2 3

2 3 6 9 [Theorem (14)]

3 6 9

2( 1) 1 2 3 [Theorem (11)]

x y z

u v w

x y z

u v w

x y z

u v w

− − −

= − − −

− − −= −

1 3 1

3 6 9

2( 1)( 1) [Theorem (11)]

1 2 3

[Theorem (15)]2( 1)( 1)

( 3 )1 2 3

x y z

u v w

x y z

u v wR r r

− − −= − −

= − −= − +

2( 1)( 1)(4) 8= − − =

48. Let 4

1 2 3

x y z

u v w =

3 1 3

[Theorem (15)] ( )

1 2 2 1 2 3

4

x y z x x y z

u v w u u v wC c c

−− =

= +

=

49. Let 4

1 2 3

x y z

u v w =

1 2 3

2 2 2

1 2 3

1 2 3

2 [Theorem (14)]

1 2 3

2( 1) 1 2 3 [Theorem (11)]

1 2 3

x y z

u v w

x y z

u v w

x y z

u v w

− − −

=− − −

= −− − −

2 3 2

2( 1)( 1) 1 2 3 [Theorem (11)]

1 2 3

[Theorem (15)]2( 1)( 1)

( )1 2 3

x y z

u v w

x y z

u v wR r r

= − − − − −

= − −= − +

2( 1)( 1)(4)

8

= − −=

50. Let 4

1 2 3

x y z

u v w =

1 3 1

2 3 2

3 6 9

3 1 3 2 3 3

1 2 3

[Theorem (15)]3 1 3 2 3 3

( =3 )1 2 3

[Theorem (15)]3 3 3

( )1 2 3

x y z

u v w

x y z

u v wR r r

x y z

u v wR r r

+ + +− − −

= − − −+

== − +

3 [Theorem (14)]

1 2 3

3(4)

12

x y z

u v w=

==

51. Solve for x:

534

3 4 5

5

5

x x

x x

x

x

=

− =− =

= −

52. Solve for x:

2

2

1 23

3 2

1 0

( 1)( 1) 0

x

x

x

x

x x

= −

− = −

− =− + =

1 0 or 1 0

1 or 1

x x

x x

− = + == = −


1076


53. Solve for x:

( ) ( ) ( )

1 1

34 2 2

51 2

3 32 4 2 41 1 25 52 1 1 2

15 4 20 2 8 3 2

11 22 11 2

11 13

1311

x

x

x

x

x

x

=−

− + =− −

− − + + + =− + =

=

=

54. Solve for x:

( ) ( ) ( )

3 2 4

51 020 1

5 51 13 2 4 02 20 01 1

3 2 5 2 2 4 1 0

6 15 4 4 0

6 7 0

6 7

76

x

x x

x

x

x

x

x

=−

− + =− −

− − − − + =− − + + =

− − =− =

= −

55. Solve for x:

( ) ( ) ( )

( )

2

2

3201 726 1

0 01 12 3 72 26 61 1

2 2 2 3 1 6 7

2 4 3 18 7

2 18 0

2 9 0

xx

x xx

x x x

x x

x x

x x

=−

− + =− −

− − − + − =

− + + − =− − =− + =

0 or 9x x= = −

56. Solve for x:

( ) ( ) ( )

( )

2

2

1 2

31 4

0 1 2

3 31 11 2 40 01 2 2 1

2 3 1 2 2 1 4

2 3 2 2 4

2 0

2 1 0

x

x x

x xx x

x x x

x x x

x x

x x

= −

− + = −

− − + = −

− − + = −

+ =+ =

1

0 or 2

x x= = −

57. Expanding the determinant:

( )

1 1

2 2

1 1 1 1

2 2 2 2

1 2 1 2 1 2 2 1

1

1 0

1

1 11 0

1 1

( ) ( ) 0

x y

x y

x y

y x x yx y

y x x y

x y y y x x x y x y

=

− + =

− − − + − =

( )( )

1 2 2 1 2 1 1 2

2 1 2 1 1 2 2 1

( )

( )

x y y y x x x y x y

y x x x y x y x y y

− + − = −− = − + −

( )

( )

2 1 1 2 1

2 1 1 2 2 1 1 2 1

2 1 1

2 1 2 1 1 2 1 2 1 1

( )

( ) ( )

( )

( )

y x x y x x

x y x y x y y y x x

x x y y

x y y x y x y y x y x

− − −= − + − − −

− −= − + − − +

( )( )

( )

2 1 1 2 1 2 1 1

2 1 1 2 1 1

2 11 1

2 1

( ) ( ) ( )

( ) ( )( )

( )( ) ( )

x x y y y y x y y x

x x y y y y x x

y yy y x x

x x

− − = − − −− − = − −

−− = −−

This is the 2-point form of the equation for a line.


1077


58. Any point ( , )x y on the line containing 2 2( , )x y

and 3 3( , )x y satisfies:

2 2

3 3

1

1 0

1

x y

x y

x y

=

If the point 1 1( , )x y is on the line containing

2 2( , )x y and 3 3( , )x y [the points are collinear],

then 1 1

2 2

3 3

1

1 0

1

x y

x y

x y

= .

Conversely, if 1 1

2 2

3 3

1

1 0

1

x y

x y

x y

= , then 1 1( , )x y is

on the line containing 2 2( , )x y and 3 3( , )x y , and the points are collinear.

59. If the vertices of a triangle are (2, 3), (5, 2), and (6, 5), then:

[ ]

[ ]

[ ]

2 5 61

3 2 52

1 1 1

2 5 3 5 3 212 5 6

1 1 1 1 1 12

12(2 5) 5(3 5) 6(3 2)

21

2( 3) 5( 2) 6(1)21

6 10 625

D =

= − +

= − − − + −

= − − − +

= − + +

=

The area of the triangle is 5 5= square units.

60. Expanding the determinant:

[ ]

2

2

2

2 22

2 2

2 2 2 2 2

2

2

1

1

1

1 11

1 1

( ) ( ) 1( )

( ) ( )( ) ( )

( )

( ) ( ) ( )

( )( )( )

x x

y y

z z

y y y yx x

z z z z

x y z x y z y z z y

x y z x y z y z yz y z

y z x xy xz yz

y z x x y z x y

y z x y x z

= − +

= − − − + −

= − − − + + −

= − − − +

= − − − −= − − −

61. If 0,a = then 0b ≠ and 0c ≠ since

0ad bc− ≠ , and the system is by s

cx dy t

=+ =

.

The solution of the system is ,s

yb

=

( )sbt dt dy tb sd

xc c bc

−− −= = = . Using Cramer’s

Rule, we get 0 b

D bcc d

= = − ,

xs b

D sd tbt d

= = − ,

00y

sD sc sc

c t= = − = − , so

xD ds tb td sdx

D bc bc− −= = =

− and

yD sc sy

D bc b−= = =−

, which is the solution. Note

that these solutions agree if 0.d =

If 0,b = then 0a ≠ and 0d ≠ since

0ad bc− ≠ , and the system is ax s

cx dy t

=+ =

.

The solution of the system is ,s

xa

=

t cx at csy

d ad− −= = . Using Cramer’s Rule, we

get 0a

D adc d

= = , 0

xs

D sdt d

= = , and


1078


ya s

D at csc t

= = − , so xD sd sx

D ad a= = =

and yD at csy

D ad−= = , which is the solution.

Note that these solutions agree if 0.c =

If 0,c = then 0a ≠ and 0d ≠ since

0ad bc− ≠ , and the system is

ax by s

dy t

+ ==

.

The solution of the system is ,t

yd

=

s by sd tbx

a ad− −= = . Using Cramer’s Rule, we

get 0

a bD ad

d= = , x

s bD sd tb

t d= = − ,

and 0ya s

D att

= = , so xD sd tbx

D ad−= = and

yD at ty

D ad d= = = , which is the solution. Note

that these solutions agree if 0.b =

If 0,d = then 0b ≠ and 0c ≠ since

0ad bc− ≠ , and the system is

ax by s

cx t

+ ==

.

The solution of the system is ,t

xc

=

s ax cs aty

b bc− −= = . Using Cramer’s Rule, we

get 00

a bD bc bc

c= = − = − ,

00x

s bD tb tb

t= = − = − , and

ya s

D at csc t

= = − , so xD tb tx

D bc c−= = =−

and

yD at cs cs aty

D bc bc− −= = =

−, which is the

solution. Note that these solutions agree if 0.a =

62. Evaluating the determinant to show the relationship:

13 12 11

23 22 21

33 32 31

22 21 23 21 23 2213 12 11

32 31 33 31 33 32

13 22 31 21 32 12 23 31 21 33

11 23 32 22 33

( ) ( )

( )

a a a

a a a

a a a

a a a a a aa a a

a a a a a a

a a a a a a a a a a

a a a a a

= − +

= − − −+ −

13 22 31 13 21 32 12 23 31 12 21 33

11 23 32 11 22 33

11 22 33 11 23 32 12 21 33 12 23 31

13 21 32 13 22 31

11 22 33 23 32 12 21 33 23 31

( ) ( )

a a a a a a a a a a a a

a a a a a a

a a a a a a a a a a a a

a a a a a a

a a a a a a a a a a

= − − ++ −

= − + + −− +

= − − + −

13 21 32 22 31

22 23 21 23 21 2211 12 13

32 33 31 33 31 32

22 23 21 23 21 2211 12 13

32 33 31 33 31 32

11 12 13

21 22 23

31 32 33

( )a a a a a

a a a a a aa a a

a a a a a a

a a a a a aa a a

a a a a a a

a a a

a a a

a a a

− −

= − + −

= − − +

= −


11 12 13

21 22 23

31 32 33

22 23 21 23 21 2211 12 13

32 33 31 33 31 32

11 22 33 23 32 12 21 33 23 31

13 21 32 22 31

11 22 33 23 32 12 21 33

( ) ( )

( )

( ) (

a a a

ka ka ka

a a a

ka ka ka ka ka kaa a a

a a a a a a

a ka a ka a a ka a ka a

a ka a ka a

ka a a a a ka a a a

= − +

= − − −+ −

= − − −

()

23 31

13 21 32 22 31

11 22 33 23 32 12 21 33 23 31

13 21 32 22 31

22 23 21 23 21 2211 12 13

32 33 31 33 31 32

11 12 13

21 22 23

)

( )

( ) ( )

( )

a

ka a a a a

k a a a a a a a a a a

a a a a a

a a a a a ak a a a

a a a a a a

a a a

k a a a

a

+ −= − − −

+ −

= − +

=

31 32 33a a

Section 11.4: Matrix Algebra

1079


64. Set up a 3 by 3 determinant in which the first column and third column are the same and evaluate:

11 12 11

21 22 21

31 32 31

22 21 21 21 21 2211 12 11

32 31 31 31 31 32

11 22 31 32 21 12 21 31 31 21

11 21 32 31 22

11 22 31 11 32 21 12 11 21 32 11 31 22

( ) ( )

( )

(0)

0

a a a

a a a

a a a

a a a a a aa a a

a a a a a a

a a a a a a a a a a

a a a a a

a a a a a a a a a a a a a

= − +

= − − −+ −

= − − + −=


11 21 12 22 13 23

21 22 23

31 32 33

22 23 21 2311 21 12 22

32 33 31 33

21 2213 23

31 32

11 21 22 33 23 32

12 22 21 33 23 31

13 23 21 32 22 31

11 22 33

( ) ( )

( )

( )( )

( )( )

( )( )

(

a ka a ka a ka

a a a

a a a

a a a aa ka a ka

a a a a

a aa ka

a a

a ka a a a a

a ka a a a a

a ka a a a a

a a a

+ + +

= + − +

+ +

= + −− + −+ + −

= 23 32 21 22 33 23 32

12 21 33 23 31 22 21 33 23 31

13 21 32 22 31 23 21 32 22 31

11 22 33 23 32 21 22 33

21 23 32 12 21 33 23 31

22 21 33 22 23 31

13 21 32

) ( )

( ) ( )

( ) ( )

( )

( )

(

a a ka a a a a

a a a a a ka a a a a

a a a a a ka a a a a

a a a a a ka a a

ka a a a a a a a

ka a a ka a a

a a a

− + −− − − −+ − + −

= − +− − −− ++ − 22 31 23 21 32

23 22 31

11 22 33 23 32 12 21 33 23 31

13 21 32 22 31

22 23 21 23 21 2211 12 13

32 33 31 33 31 32

11 12 13

21 22 23

31 32 33

)

( ) ( )

( )

a a ka a a

ka a a

a a a a a a a a a a

a a a a a

a a a a a aa a a

a a a a a a

a a a

a a a

a a a

+−

= − − −+ −

= − +

=

Section 11.4

1. square

2. True

3. columns; rows

4. False

5. inverse

6. singular

7. True

8. 1A B−

9. 0 3 5 4 1 0

1 2 6 2 3 2

0 4 3 1 5 0

1 ( 2) 2 3 6 ( 2)

4 4 5

1 5 4

A B−

+ = +− −

+ + − +=

+ − + + −−

=−

10. 0 3 5 4 1 0

1 2 6 2 3 2

0 4 3 1 5 0

1 ( 2) 2 3 6 ( 2)

4 2 5

3 1 8

A B−

− = −− −

− − − −=

− − − − −− −

=−

11. 0 3 5

4 41 2 6

4 0 4 3 4( 5)

4 1 4 2 4 6

0 12 20

4 8 24

A−

=

⋅ ⋅ −=

⋅ ⋅ ⋅−

=

12. 4 1 0

3 32 3 2

3 4 3 1 3 03 2 3 3 3 2

12 3 06 9 6

B− = −− −

− ⋅ − ⋅ − ⋅=

− ⋅ − − ⋅ − ⋅ −− −

=−


1080


13. 0 3 5 4 1 0

3 2 3 21 2 6 2 3 2

0 9 15 8 2 03 6 18 4 6 4

8 7 157 0 22

A B−

− = −− −

−= −

− −− −

=

14. 0 3 5 4 1 0

2 4 2 41 2 6 2 3 2

0 6 10 16 4 0

2 4 12 8 12 8

16 10 10

6 16 4

A B−

+ = +− −

−= +

− −−

=−

15. 4 1

0 3 56 2

1 2 62 3

0(4) 3(6) ( 5)( 2) 0(1) 3(2) ( 5)(3)

1(4) 2(6) 6( 2) 1(1) 2(2) 6(3)

28 9

4 23

AC−

= ⋅−

+ + − − + + −=

+ + − + +−

=

16. 4 1

4 1 06 2

2 3 22 3

4(4) 1(6) 0( 2) 4(1) 1(2) 0(3)

2(4) 3(6) ( 2)( 2) 2(1) 3(2) ( 2)(3)

22 6

14 2

BC = ⋅− −

−+ + − + +

=− + + − − − + + −

=−

17. 4 1

0 3 56 2

1 2 62 3

4(0) 1(1) 4(3) 1(2) 4( 5) 1(6)

6(0) 2(1) 6(3) 2(2) 6( 5) 2(6)

2(0) 3(1) 2(3) 3(2) 2( 5) 3(6)

1 14 14

2 22 18

3 0 28

CA−

= ⋅−

+ + − += + + − +

− + − + − − +−

= −

18. 4 1

4 1 06 2

2 3 22 3

4(4) 1( 2) 4(1) 1(3) 4(0) 1( 2)

6(4) 2( 2) 6(1) 2(3) 6(0) 2( 2)

2(4) 3( 2) 2(1) 3(3) 2(0) 3( 2)

14 7 2

20 12 4

14 7 6

CB = ⋅− −

−+ − + + −

= + − + + −− + − − + − + −

−= −

− −

19. 4 1

0 3 5 4 1 0( ) 6 2

1 2 6 2 3 22 3

4 14 4 5

6 21 5 4

2 3

15 21 16

22 34 22

11 7 22

C A B−

+ = +− −

−

−= ⋅

−−

−= −

−

20. 4 1

0 3 5 4 1 0( ) 6 2

1 2 6 2 3 22 3

4 14 4 5

6 21 5 4

2 3

50 3

18 21

A B C−

+ = +− −

−

−= ⋅

−−

−=

21. 2

4 10 3 5 1 0

3 6 2 31 2 6 0 1

2 3

28 9 3 0

4 23 0 3

25 9

4 20

AC I−

− = ⋅ −−

−= −

−=


1081


22. 3

4 1 1 0 00 3 5

5 6 2 5 0 1 01 2 6

2 3 0 0 1

1 14 14 5 0 0

2 22 18 0 5 0

3 0 28 0 0 5

6 14 14

2 27 18

3 0 33

CA I−

+ = ⋅ +−

−= − +

−= −

23. CA CB−

4 1 4 1

0 3 5 4 1 06 2 6 2

1 2 6 2 3 22 3 2 3

1 14 14 14 7 2

2 22 18 20 12 4

3 0 28 14 7 6

13 7 12

18 10 14

17 7 34

−= ⋅ − ⋅

− −− −

− −= − − −

− −− −

= − −−

24. 4 1 4 1

0 3 5 4 1 06 2 6 2

1 2 6 2 3 22 3 2 3

28 9 22 6

4 23 14 2

50 3

18 21

AC BC+

−= ⋅ + ⋅

− −− −

−= +

−−

=

25. 11

12

13

14

21

22

23

24

2(2) ( 2)(3) 2

2(1) ( 2)( 1) 4

2(4) ( 2)(3) 2

2(6) ( 2)(2) 8

1(2) 0(3) 2

1(1) 0( 1) 1

1(4) 0(3) 4

1(6) 0(2) 6

a

a

a

a

a

a

a

a

= + − = −= + − − == + − == + − == + == + − == + == + =

2 2 2 1 4 6 2 4 2 8

1 0 3 1 3 2 2 1 4 6

− −=

−

26. 11

12

13

14

21

22

23

24

4( 6) 1(2) 22

4(6) 1(5) 29

4(1) 1(4) 8

4(0) 1( 1) 1

2( 6) 1(2) 10

2(6) 1(5) 17

2(1) 1(4) 6

2(0) 1( 1) 1

a

a

a

a

a

a

a

a

= − + = −= + == + == + − = −= − + = −= + == + == + − = −

4 1 6 6 1 0 22 29 8 1

2 1 2 5 4 1 10 17 6 1

− − −=

− − −

27. 1 2

1 2 31 0

0 1 42 4

1(1) 2( 1) 3(2) 1(2) 2(0) 3(4)

0(1) ( 1)( 1) 4(2) 0(2) ( 1)(0) 4(4)

5 14

9 16

−−

+ − + + +=

+ − − + + − +

=

28. 1 1

2 8 13 2

3 6 00 5

1(2) ( 1)(3) 1(8) ( 1)(6) 1( 1) ( 1)(0)

( 3)(2) 2(3) ( 3)(8) 2(6) ( 3)( 1) 2(0)

0(2) 5(3) 0(8) 5(6) 0( 1) 5(0)

−−

−

+ − + − − + −= − + − + − − +

+ + − +

1 2 1

0 12 3

15 30 0

− −= −

29. 1 0 1 1 3

2 4 1 6 2

3 6 1 8 1

1(1) 0(6) 1(8) 1(3) 0(2) 1( 1)

2(1) 4(6) 1(8) 2(3) 4(2) 1( 1)

3(1) 6(6) 1(8) 3(3) 6(2) 1( 1)

9 2

34 13

47 20

−+ + + + −

= + + + + −+ + + + −

=


1082


30. 4 2 3 2 6

0 1 2 1 1

1 0 1 0 2

4(2) ( 2)(1) 3(0) 4(6) ( 2)( 1) 3(2)

0(2) 1(1) 2(0) 0(6) 1( 1) 2(2)

1(2) 0(1) 1(0) 1(6) 0( 1) 1(2)

−−

−+ − + + − − +

= + + + − +− + + − + − +

6 32

1 3

2 4

=− −

31. 2 1

1 1A =

Augment the matrix with the identity and use row operations to find the inverse:

( )1 2

2 1 2

2 1 1 0

1 1 0 1

Interchange1 1 0 1

and 2 1 1 0

1 1 0 1 2

0 1 1 2

r r

R r r

→

→ = − +− −

( )

( )

2 2

1 2 1

1 1 0 1

0 1 1 2

1 0 1 1

0 1 1 2

R r

R r r

→ = −−

−→ = − +

−

Thus, 1 1 1

1 2A− −

=−

.

32. 3 1

2 1A

−=

−


( )

( )

1 2 1

2 1 2

3 1 1 0

2 1 0 1

1 0 1 1

2 1 0 1

1 0 1 1 2

0 1 2 3

R r r

R r r

−−

→ = +−

→ = +

Thus, 1 1 1

2 3A− = .

33. 6 5

2 2A =


( )

( )

1 2

2 1 2

111 122

2 2

52

1 2 1

6 5 1 0

2 2 0 1

Interchange2 2 0 1

and 6 5 1 0

2 2 0 1 3

0 1 1 3

1 1 0

0 1 1 3

1 0 1

0 1 1 3

r r

R r r

R r

R r

R r r

→

→ = − +− −

=→

− = −

−→ = − +−

Thus, 5

1 21

1 3A− −=

−.

34. 4 1

6 2A

−=

−


( )

( )

( )

( )

1 114 4

1 14

1 14 4

2 1 2312 2

1 14 4

2 2

112

1 24

4 1 1 0

6 2 0 1

1 0

6 2 0 1

1 0 6

0 1

1 0 2

0 1 3 2

1 0 1

0 1 3 2

R r

R r r

R r

R r r

−−

− −→ = −−

− −→ = − +

−

− −→ = −− −

− −→ = +− −

Thus, 1

1 21

3 2A− − −=

− −.


1083


35. 2 1

where 0.A aa a

= ≠


( )

( )

( )

( )

1 112 2

1 12

1 12 2

2 1 21 12 2

1 122 2

2 22

11

1 2 122

2 1 1 0

0 1

1 0

0 1

1 0

0 1

1 0

0 1 1

1 0 1

0 1 1

aa

a

a

a a

R ra a

R a r ra a

R r

R r r

→ =

→ = − +−

→ =−

−→ = − +

−

Thus, 1

12

1

1a

a

A− −=

−.

36. 3

where 0.2

bA b

b= ≠


( )

( )

( )

( )

2 1 2

3 11

1 1

3 1

2 2

323

1 2 1

3 1 0

2 0 1

3 1 0

0 1 1 1

1 0

0 1 1 1

1 0

0 1 1 1

1 0

0 1 1 1

b bb

b b

b bb

b

b

bR r r

R r

R r

R r r

→ = − +− −

→ =− −

→ = −−

−→ = − +

−

Thus, 32

1

1 1b bA− −

=−

.

37. 1 1 1

0 2 1

2 3 0

A

−= −

− −


( )

( )

3 1 3

1 1 12 22 2 2

1 12 2

1 2 11 12 2

3 2 3512 2

1 12 2

1 1 1 1 0 0

0 2 1 0 1 0

2 3 0 0 0 1

1 1 1 1 0 0

0 2 1 0 1 0 2

0 5 2 2 0 1

1 1 1 1 0 0

0 1 0 0

0 5 2 2 0 1

1 0 1 0

0 1 0 0 5

0 0 2 1

1 0 1

R r r

R r

R r r

R r r

−−

− −−

→ − = +−−

→ − − = −−

−= +

→ − −= +

− −

−→ ( )1 1

3 32 2

11 3 12

12 3 22

0

0 1 0 0 2

0 0 1 4 5 2

1 0 0 3 3 1

0 1 0 2 2 1

0 0 1 4 5 2

R r

R r r

R r r

− − = −− −

− = − +→ − −

= +− −

Thus, 1

3 3 1

2 2 1

4 5 2

A−−

= − −− −

.

38. 1 0 2

1 2 3

1 1 0

A = −−


( )

2 1 2

3 1 3

5 1 1 12 22 2 2 2

1 0 2 1 0 0

1 2 3 0 1 0

1 1 0 0 0 1

1 0 2 1 0 0

0 2 5 1 1 0

0 1 2 1 0 1

1 0 2 1 0 0

0 1 0

0 1 2 1 0 1

R r r

R r r

R r

−−

= +→

= − +− − −

→ =− − −


1084


( )5 1 13 2 32 2 2

1 1 12 2 2

1 0 2 1 0 0

0 1 0

0 0 1

R r r→ = +−

( )5 1 13 32 2 2

1 3 1

52 3 22

1 0 2 1 0 0

0 1 0 2

0 0 1 1 1 2

1 0 0 3 2 4 20 1 0 3 2 5

0 0 1 1 1 2

R r

R r r

R r r

→ =−

− − = − +→ − −

= − +−

Thus, 1

3 2 4

3 2 5

1 1 2

A−− −

= − −−

.

39. 1 1 1

3 2 1

3 1 2

A = −


1 1 1 1 0 0

3 2 1 0 1 0

3 1 2 0 0 1

−

2 1 2

3 1 3

1 1 1 1 0 03

0 1 4 3 1 0 3

0 2 1 3 0 1

R r r

R r r

= − +→ − − −

= − +− − −

( )2 2

1 1 1 1 0 0

0 1 4 3 1 0

0 2 1 3 0 1

R r→ − = −− − −

( )13 37

3 2 17 7 7

5 317 7 7

1 3 19 1 47 7 7

2 3 23 2 17 7 7

1 0 3 2 1 0

0 1 4 3 1 0

0 0 1

1 0 03

0 1 0 4

0 0 1

R r

R r r

R r r

− −→ − =

−

−= +

→ −= − +

−

Thus,

5 317 7 7

1 9 1 47 7 73 2 17 7 7

A−

−= −

−.

40. 3 3 1

1 2 1

2 1 1

A =−


3 3 1 1 0 0

1 2 1 0 1 0

2 1 1 0 0 1−

( )

1 2

2 1 2

3 1 3

2 1 12 23 3 3

1 23 32 13 37 53 3

1 2 1 0 1 0Interchange

3 3 1 1 0 0 and

2 1 1 0 0 1

1 2 1 0 1 03

0 3 2 1 3 0 2

0 5 1 0 2 1

1 2 1 0 1 0

0 1 1 0

0 5 1 0 2 1

1 0 1 0

0 1 1 0

0 0 3 1

r r

R r r

R r r

R r

→−

= − +→ − − −

= − +− − −

→ − = −

− − −

− −

→ −

−

( )

1 2 1

3 2 3

1 23 3

32 13 33 3 7

5 9 37 7 7

3 4 17 7 7 1

1 3 131 1 27 7 7 2

2 3 235 9 37 7 7

2

5

1 0 1 0

0 1 1 0

0 0 1

1 0 0

0 1 0

0 0 1

R r r

R r r

R r

R r r

R r r

= − += +

− −

→ − =

−

−= +

→ −= − +

−

Thus,

3 4 17 7 7

1 1 1 27 7 75 9 37 7 7

A−−

= −

−

.


1085


41. 2 8

5

x y

x y

+ =+ =

Rewrite the system of equations in matrix form: 2 1 8

, ,1 1 5

xA X B

y= = =

Find the inverse of 1 and solve A X A B−= :

From Problem 29, 1 1 1

1 2A− −

=−

, so

1 1 1 8 3

1 2 5 2X A B− −

= = =−

.

The solution is 3, 2x y= = or (3, 2) .

42. 3 8

2 4

x y

x y

− =− + =


, ,2 1 4

xA X B

y

−= = =

−



2 3A− = , so

1 1 1 8 12

2 3 4 28X A B−= = = .


43. 2 0

5

x y

x y

+ =+ =


, ,1 1 5

xA X B

y= = =



1 2A− −

=−

, so

1 1 1 0 5

1 2 5 10X A B− − −

= = =−

.

The solution is 5, 10x y= − = or ( 5,10)− .

44. 3 4

2 5

x y

x y

− =− + =


, ,2 1 5

xA X B

y

−= = =

−



2 3A− = , so

1 1 1 4 9

2 3 5 23X A B−= = = .


45. 6 5 7

2 2 2

x y

x y

+ =+ =


, ,2 2 2

xA X B

y= = =


From Problem 31, 5

1 21

1 3A− −=

−, so

51 2

7 212 11 3

X A B− −= = =−−

.


46. 4 0

6 2 14

x y

x y

− + =− =


, ,6 2 14

xA X B

y

−= = =

−


From Problem 32, 1

1 21

3 2A− − −=

− −, so

11 2

0 7114 283 2

X A B− −− −= = =−− −

.

The solution is 7, 28x y= − = − or ( 7, 28)− − .


1086


47. 6 5 13

2 2 5

x y

x y

+ =+ =


, ,2 2 5

xA X B

y= = =


From Problem 31, 5

1 21

1 3A− −=

−, so

5 11 22

1315 21 3

X A B− −= = =−

.

The solution is 1

, 22

x y= = or 1

, 22

.

48. 4 5

6 2 9

x y

x y

− + =− = −


, ,6 2 9

xA X B

y

−= = =

− −


From Problem 32, 1

1 21

3 2A− − −=

− −, so

1 11 2 2

5193 2 3

X A B− − − −= = =−− −

.

The solution is 1

, 32

x y= − = or 1

, 32

− .

49. 2 3

0x y

aax ay a

+ = −≠

+ = −


, ,x

A X Ba a y a

−= = =

−


From Problem 33, 1

12

1

1a

a

A− −=

−, so

11

2

1 3 2

11a

a

X A Ba

− − − −= = =

−−.

The solution is 2, 1x y= − = or ( 2, 1)− .

50. 3 2 3

02 2 2

bx y bb

bx y b

+ = +≠

+ = +


, ,2 2 2

b x bA X B

b y b

+= = =

+


From Problem 34, 32

1

1 1b bA− −

=−

, so

321 2 3 2

2 2 11 1b b b

X A Bb

− − += = =

+−.

The solution is 2, 1x y= = or (2, 1).

51. 7

20

5

x yaa

ax ay

+ =≠

+ =

Rewrite the system of equations in matrix form: 72 1

, ,5ax

A X Ba a y

= = =


From Problem 33, 1

12

1

1a

a

A− −=

−, so

1 271

32

1

1 5a aa

a a

X A B− −= = =

−.

The solution is 2 3

,x ya a

= = or 2 3

,a a

.

52. 3 14

02 10

bx yb

bx y

+ =≠

+ =

Rewrite the system of equations in matrix form: 3 14

, ,2 10

b xA X B

b y= = =


From Problem 34, 32

1

1 1b bA− −

=−

, so

3 221 14

10 41 1bb bX A B− −

= = =−

.

The solution is 2

, 4x yb

= = or 2

, 4b

.


1087


53. 0

2 1

2 3 5

x y z

y z

x y

− + =− + = −

− − = −

Rewrite the system of equations in matrix form: 1 1 1 0

0 2 1 , , 1

2 3 0 5

x

A X y B

z

−= − = = −

− − −


From Problem 35, 1

3 3 1

2 2 1

4 5 2

A−−

= − −− −

, so

1

3 3 1 0 2

2 2 1 1 3

4 5 2 5 5

X A B−− −

= = − − − =− − −

.

The solution is 2, 3, 5x y z= − = = or ( 2, 3, 5)− .

54. 2 6

2 3 5

6

x z

x y z

x y

+ =− + + = −

− =


1 2 3 , , 5

1 1 0 6

x

A X y B

z

= − = = −−


From Problem 36, 1

3 2 4

3 2 5

1 1 2

A−− −

= − −−

, so

1

3 2 4 6 4

3 2 5 5 2

1 1 2 6 1

X A B−− −

= = − − − = −−

.


55. 2

2 2

12 3

2

x y z

y z

x y

− + =− + =

− − =

Rewrite the system of equations in matrix form:

12

1 1 1 2

0 2 1 , , 2

2 3 0

x

A X y B

z

−= − = =

− −


From Problem 35, 1

3 3 1

2 2 1

4 5 2

A−−

= − −− −

, so

12

1 12

12

3 3 1 2

2 2 1 2

4 5 2 1

X A B−−

= = − − = −− −

The solution is 1 1

, , 12 2

x y z= = − = or

1 1, , 1

2 2− .

56.

2 2

32 3

22

x z

x y z

x y

+ =

− + + = −

− =


32

1 0 2 2

1 2 3 , ,

1 1 0 2

x

A X y B

z

= − = = −−


From Problem 36, 1

3 2 4

3 2 5

1 1 2

A−− −

= − −−

, so

1 32

12

3 2 4 2 1

3 2 5 1

1 1 2 2

X A B−− −

= = − − − = −−

.

The solution is 1

1, 1,2

x y z= = − = or 1

1, 1,2

− .


1088


57. 9

3 2 8

3 2 1

x y z

x y z

x y z

+ + =+ − =+ + =


3 2 1 , , 8

3 1 2 1

x

A X y B

z

= − = =


From Problem 37,

5 317 7 7

1 9 1 47 7 73 2 17 7 7

A−

−= −

−, so

5 3 3417 7 7 7

1 9 851 47 7 7 73 2 1 127 7 7 7

9

8

1

X A B−

− −= = − =

−.


, ,7 7 7

x y z= − = = or

34 85 12, ,

7 7 7− .

58. 3 3 8

2 5

2 4

x y z

x y z

x y z

+ + =+ + =− + =


1 2 1 , , 5

2 1 1 4

x

A X y B

z

= = =−


From Problem 38,

3 4 17 7 7

1 1 1 27 7 75 9 37 7 7

A−

−= −

−, so

3 84 17 7 7 7

1 51 1 27 7 7 75 9 3 177 7 7 7

8

5

4

X A B−

−= = − =

−.


, ,7 7 7

x y z= = = or

8 5 17, ,

7 7 7.

59.

2

73 2

310

3 23

x y z

x y z

x y z

+ + =

+ − =

+ + =


73

103

1 1 1 2

3 2 1 , ,

3 1 2

x

A X y B

z

= − = =


From Problem 37,

5 317 7 7

1 9 1 47 7 73 2 17 7 7

A−

−= −

−, so

5 31 17 7 7 3

1 9 71 47 7 7 3

23 102 137 7 7 3

2

1X A B−

−= = − =

−.

The solution is 1 2

, 1,3 3

x y z= = = or 1 2

, 1,3 3

.

60. 3 3 1

2 0

2 4

x y z

x y z

x y z

+ + =+ + =− + =


1 2 1 , , 0

2 1 1 4

x

A X y B

z

= = =−


From Problem 38,

3 4 17 7 7

1 1 1 27 7 75 9 37 7 7

A−

−= −

−, so

3 4 17 7 7

1 1 1 27 7 75 9 37 7 7

1 1

0 1

4 1

X A B−

−= = − = −

−.



1089


61. 4 2

2 1A =


( )

( )

12 1 21 2

2

1 112 4

1 11 42

4 2 1 0

2 1 0 1

4 2 1 0

0 0 1

1 0

0 0 1

R r r

R r

→ = − +−

→ =−

There is no way to obtain the identity matrix on the left. Thus, this matrix has no inverse.

62. 123

6 1A

−=−


( )

( )

12

12

2 1 2

1 116 3

1 13

3 1 0

6 1 0 1

3 1 0 2

0 0 2 1

1 0

0 0 2 1

R r r

R r

−−

−→ = +

− −→ = −

There is no way to obtain the identity matrix on the left. Thus, this matrix has no inverse.

63. 15 3

10 2A =


( )

( )

22 1 22 3

3

1 15 15 1

1 11523

15 3 1 0

10 2 0 1

15 3 1 0

0 0 1

1 0

0 0 1

R r r

R r

→ = − +−

→ =−

There is no way to obtain the identity matrix on the left; thus, there is no inverse.

64. 3 0

4 0A

−=


( )

( )

42 1 24 3

3

13 1

1 1343

3 0 1 0

4 0 0 1

3 0 1 0

0 0 1

1 0 0

0 0 1

R r r

R r

−

−→ = +

−→ = −


65. 3 1 1

1 4 7

1 2 5

A

− −= − −


3 1 1 1 0 0

1 4 7 0 1 0

1 2 5 0 0 1

− −− −

( )

1 3

2 1 2

3 1 3

1 1 12 26 6 6


1 4 7 0 1 0 and

3 1 1 1 0 0

1 2 5 0 0 1

0 6 12 0 1 1 3

0 7 14 1 0 3

1 2 5 0 0 1

0 1 2 0

0 7 14 1 0 3

r r

R r r

R r r

R r

→ − −− −

= − +→ − − −

= +

→ − = −

1 23 3

1 2 11 16 6

3 2 37 116 6

1 0 1 02

0 1 2 0 7

0 0 0 1

R r r

R r r

= − +→ −

= − +



1090


66. 1 1 3

2 4 1

5 7 1

A

−= −

−


1 1 3 1 0 0

2 4 1 0 1 0

5 7 1 0 0 1

−−

−

( )

2 1 2

3 1 3

7 1 1 12 26 3 6 6

11 2 16 3 6

1 2 17 1 16 3 6

3 2 3

1 1 3 1 0 02

0 6 7 2 1 0 5

0 12 14 5 0 1

1 1 3 1 0 0

0 1 0

0 12 14 5 0 1

1 0 0

0 1 0 12

0 0 0 1 2 1

R r r

R r r

R r

R r r

R r r

−= − +

→ − −= +

−−

→ − − = −−

−= − +

→ − −= − +


67. 25 61 12

18 2 4

8 35 21

A

−= −

Thus, 10.01 0.05 0.01

0.01 0.02 0.01

0.02 0.01 0.03

A−−

≈ −−

68. 18 3 4

6 20 14

10 25 15

A

−= −

−

Thus, 10.26 0.29 0.20

1.21 1.63 1.20

1.84 2.53 1.80

A−− −

≈ −−

69.

44 21 18 6

2 10 15 5

21 12 12 4

8 16 4 9

A−

=−

− −

Thus, 1

0.02 0.04 0.01 0.01

0.02 0.05 0.03 0.03

0.02 0.01 0.04 0.00

0.02 0.06 0.07 0.06

A−

− −− −

≈−

−

.

70.

16 22 3 5

21 17 4 8

2 8 27 20

5 15 3 10

A

−−

=

− −

Thus, 1

0.01 0.04 0.00 0.03

0.02 0.02 0.01 0.01

0.04 0.02 0.04 0.06

0.05 0.02 0.00 0.09

A− −=

−− −

.

71. 25 61 12 10

18 12 7 ; 9

3 4 1 12

A B

−= − = −

−

Enter the matrices into a graphing utility and use 1A B− to solve the system. The result is shown

below:

Thus, the solution to the system is 4.57x ≈ ,

6.44y ≈ − , 24.07z ≈ − or (4.57, 6.44, 24.07)− − .


1091


72. 25 61 12 15

18 12 7 ; 3

3 4 1 12

A B

−= − = −

−

Enter the matrices into a graphing utility and use 1A B− to solve the system. The result is shown

below:

Thus, the solution to the system is 4.56x ≈ ,

6.06y ≈ − , 22.55z ≈ − or (4.56, 6.06, 22.55)− − .

73. 25 61 12 21

18 12 7 ; 7

3 4 1 2

A B

−= − =

− −

Thus, the solution to the system is 1.19x ≈ − ,

2.46y ≈ , 8.27z ≈ or ( 1.19, 2.46, 8.27)− .

74. 25 61 12 25

18 12 7 ; 10

3 4 1 4

A B

−= − =

− −

Thus, the solution to the system is 2.05x ≈ − ,

3.88y ≈ , 13.36z ≈ or ( 2.05, 3.88, 13.36)− .

75. 2 3 11

5 7 24

x y

x y

+ =+ =

Multiply each side of the first equation by 5, and each side of the second equation by 2− . Then add the equations to eliminate x:

10 15 55

10 14 48

x y

x y

+ =− − = −

7y = Substitute and solve for x:

( )2 3 7 11

2 21 11

2 10

5

x

x

x

x

+ =+ =

= −= −

The solution of the system is 5, 7x y= − = or

using ordered pairs ( )5,7− .

76. 2 8 8

7 13

x y

x y

+ = −+ = −

Multiply each side of the second equation by 2− . Then add the equations to eliminate x: 2 8 8

2 14 26

x y

x y

+ = −− − =

6 18

3

y

y

− == −

Substitute and solve for x: ( )7 3 13

21 13

8

x

x

x

+ − = −− = −

=


using ordered pairs ( )8, 3− .

77. 2 4 2

3 5 2 17

4 3 22

x y z

x y z

x y

− + =− + − =

− = −


1 2 4 2

3 5 172

3 04 22

−− −

− −

2 1 2

3 1 3

1 2 4 20 10 231 30 5 16 30 4

R r rR r r

−− = +

− − = − +

2 2

1 2 4 20 10 2310 5 16 30

R r−

− − = −− −

3 2 3

1 2 4 20 10 2310 0 34 85 5R r r

−− −

= − +

52 3 3

1 2 4 20 10 231

/ 340 0 1 R r

−− −

=

1 3 1

2 3 252

0 81 2 40 01 2 100 0 1

R r rR r r

−− = − += +


1092


1 2 1

52

0 01 4 20 01 20 0 1

R r r− = +

The solution is 5

4, 2,2

x y z= − = = or 5

4,2,2

− .

78. 2 3 2

4 3 6

6 2 2

x y z

x z

y z

+ − = −+ =

− =


32 1 2

0 3 64

0 6 2 2

− −

−

3 12 2 1 11 1 / 2

0 3 64

0 6 2 2

R r− − =

−

3 12 2

2 1 2

1 1

0 6 5 10 4

0 6 2 2

R r r

− −− = − +

−

( )

3 12 2

5 52 26 3

1 1

/ 60 1

0 6 2 2

R r

− −− − = −

−

3 12 2

5 56 3

3 2 3

1 1

0 160 0 3 12 R r r

− −− −

= − +

3 12 2

5 56 3

3 3

1 1

0 1/30 0 1 4 R r

− −− −

=

3 12 1 3 125

53 2 3 26

01 1

0 01

0 0 1 4

R r r

R r r

= +

= +

3 32 1 2 1253

0 01

0 01

0 0 1 4

R r r− = − +

The solution is 3 5

, , 42 3

x y z= − = = or

3 5, , 4

2 3− .

79. 5 4 2

5 4 3

7 13 4 17

x y z

x y z

x y z

− + =− + − =

+ − =


5 1 4 2

5 31 4

7 13 174

−− −

−

1 2

2 1

5 31 4

5 1 4 2

7 13 174

R r

R r

− − = −− =

−

2 1 2

3 1 3

5 31 4

0 16 1724 5

0 48 32 38 7

R r r

R r r

− −− = − +− = − +

2 172 23 24

5 31 4

/ 240 1

0 48 32 38

R r

− −− =

−

2 173 24

3 2 3

5 31 4

0 1480 0 0 4 R r r

− −−

= − +

The last row of our matrix is a contradiction. Therefore, the system is inconsistent. The solution set is { } , or ∅ .

80. 3 2 2

2 6 7

2 2 14 17

x y z

x y z

x y z

+ − =+ + = −

+ − =


3 2 1 2

6 72 1

172 2 14

−−

−

1 1 27 91 1

6 72 1

172 2 14

R r r− = −−

−

2 1 2

3 1 3

7 91 1

0 20 251 2

0 0 0 1 2

R r r

R r r

−−− = − +

− = − +

The last row of our matrix is a contradiction. Therefore, the system is inconsistent. The solution set is { } , or ∅ .


1093


81. 2 3 4

3 2 3

5 6

x y z

x y z

y z

− + =− + − = −

− + =


32 1 4

3 32 1

0 5 61

−− −−

−

3 12 2 1 11 2 / 2

3 32 1

0 5 61

R r− =− −−

−

3 12 25 1

2 1 22 2

1 2

30 3

0 5 61

R r r

−

− = +

−

3 12 2

61 25 5 2 25

1 2

0 1

0 5 61

R r

−

− − = −−

3 12 2

615 5

3 2 3

1 2

0 150 0 0 0 R r r

−

− −= +

1 1 35 5 1 2 1261

5 5

01

0 1

0 0 0 0

R r r= +− −

Since the last row yields an identity, and no contradictions exist in the other rows, there are an infinite number of solutions. The solution is

1 15 5

x z= − + , 1 65 5

y z= − , and z is any real

number. That is,

( )

}

1 1 1 6, , | , ,

5 5 5 5

is any real number

x y z x z y z

z

= − + = −

82. 4 3 2 6

3 2

9 6

x y z

x y z

x y z

− + + =+ − = −+ + =


3 64 2

3 1 1 2

9 61 1

−− −

1 3

3 1

9 61 1

3 1 1 2

3 64 2

R r

R r

=− −

− =

2 1 2

3 1 3

9 61 1

0 26 204 3

0 39 6 30 4

R r r

R r r

− −− = − += +

( )1022 213 13

9 61 1

/ 260 1

0 39 6 30

R r= −

10213 13

3 2 3

9 61 1

0 1390 0 0 0 R r r= − +

5 1213 13 1 2 1

10213 13

01 9

0 1

0 0 0 0

R r r− − = − +

Since the last row yields an identity, and no contradictions exist in the other rows, there are an infinite number of solutions. The solution is

5 1213 13

x z= − , 2 10

13 13y z= − + , and z is any real

number. That is,

( )

}

5 12 2 10, , | , ,

13 13 13 13

is any real number

x y z x z y z

z

= − = − +

83. a. 6 9 80.00

; 3 12 277.80

A B= =

b. 6 9 80.00

3 12 277.80

6(80.00) 9(277.80) 2980.20

3(80.00) 12(277.80) 3573.60

AB =

+= =

+

Nikki’s total tuition is $2980.20, and Joe’s total tuition is $3573.60.

84. a. 4000 3000 0.011

; 2500 3800 0.006

A B= =

b. 4000 3000 0.011

2500 3800 0.006

4000(0.011) 3000(0.006) 62.00

2500(0.011) 3800(0.006) 50.30

AB =

+= =

+

After one month, Jamal’s loans accrued


1094


$62.00 in interest, and Stephanie’s loans accrued $50.30 in interest.

c. 4000 3000 1 0.011

( )2500 3800 1 0.006

4000 3000 1.011

2500 3800 1.006

4000(1.011) 3000(1.006)

2500(1.011) 3800(1.006)

7062.00

6350.30

A C B+ = +

=

+=

+

=

Jamal’s loan balance after one month was $7062.00, and Stephanie’s loan balance was $6350.30.

85. a. The rows of the 2 by 3 matrix represent stainless steel and aluminum. The columns represent 10-gallon, 5-gallon, and 1-gallon.

The 2 by 3 matrix is: 500 350 400

700 500 850.

The 3 by 2 matrix is:

500 700

350 500

400 850

.

b. The 3 by 1 matrix representing the amount of

material is:

15

8

3

.

c. The days usage of materials is: 15

500 350 400 11,5008

700 500 850 17,0503

⋅ =

Thus, 11,500 pounds of stainless steel and 17,050 pounds of aluminum were used that day.

d. The 1 by 2 matrix representing cost is: [ ]0.10 0.05 .

e. The total cost of the day’s production was:

[ ] [ ]11,5000.10 0.05 2002.50

17,050⋅ = .

The total cost of the day’s production was $2002.50.

86. a. The rows of the 2 by 3 matrix represent the location. The columns represent the type of car sold. The 2 by 3 matrix for January is:

400 250 50

450 200 140. The 2 by 3 matrix for

February is: 350 100 30

350 300 100.

b. Adding the matrices: 400 250 50 350 100 30

450 200 140 350 300 100

750 350 80

800 500 240

+

=

c. The 3 by 1 matrix representing profit:

100

150

200

.

d. Multiplying to find the profit at each location: 100

750 350 80 143,500150

800 500 240 203,000200

⋅ = .

The city location has a two-month profit of $143,500. The suburban location has a two-month profit of $203,000.

87. a. 2 1 11 1 01 1 1

K =


2 1 1 1 0 01 1 0 0 1 01 1 1 0 0 1

1 2

2 1 2

3 1 3


2 1 1 1 0 0 and

1 1 1 0 0 1

1 1 0 0 1 02

0 1 1 1 2 0 0 0 1 0 1 1

r r

R r r

R r r

→

= − +→ − −

= − +−


1095


( )

( )

( )

2 2

2 2 3

1 1 2

1 1 0 0 1 0

0 1 1 1 2 0

0 0 1 0 1 1

1 1 0 0 1 0

0 1 0 1 1 1

0 0 1 0 1 1

1 0 0 1 0 1

0 1 0 1 1 1

0 0 1 0 1 1

R r

R r r

R r r

→ − − = −−

→ − = +−

−→ − = −

−

Thus, 1

1 0 1

1 1 1

0 1 1

K −−

= −−

.

b. 1

47 34 33 1 0 1

44 36 27 1 1 1

47 41 20 0 1 1

M E K −−

= ⋅ = −−

13 1 20

8 9 19

6 21 14

=

because 11

12

13

21

22

23

31

32

33

47(1) 34( 1) 33(0) 13

47(0) 34(1) 33( 1) 1

47( 1) 34(1) 33(1) 20

44(1) 36( 1) 27(0) 8

44(0) 36(1) 27( 1) 9

44( 1) 36(1) 27(1) 19

47(1) 41( 1) 20(0) 6

47(0) 41(1) 20( 1) 21

a

a

a

a

a

a

a

a

a

= + − + == + + − == − + + == + − + == + + − == − + + == + − + == + + − == 47( 1) 41(1) 20(1) 14− + + =

c. 13 ;1 ; 20 ; 8 ; 9 ;

19 ;6 ; 21 ;14

M A T H I

S F U N

→ → → → →→ → → →

The message: Math is fun.

88. 2

0.4 0.2 0.1 0.4 0.2 0.1

0.5 0.6 0.5 0.5 0.6 0.5

0.1 0.2 0.4 0.1 0.2 0.4

0.27 0.22 0.18

0.55 0.56 0.55

0.18 0.22 0.27

P =

=

because

11

12

13

21

22

23

0.4(0.4) 0.2(0.5) 0.1(0.1) 0.27

0.4(0.2) 0.2(0.6) 0.1(0.2) 0.22

0.4(0.1) 0.2(0.5) 0.1(0.4) 0.18

0.5(0.4) 0.6(0.5) 0.5(0.1) 0.55

0.5(0.2) 0.6(0.6) 0.5(0.2) 0.56

0.5(0.1) 0.6(0.5) 0.5

a

a

a

a

a

a

= + + == + + == + + == + + == + + == + +

31

32

33

(0.4) 0.55

0.1(0.4) 0.2(0.5) 0.4(0.1) 0.18

0.1(0.2) 0.2(0.6) 0.4(0.2) 0.22

0.1(0.1) 0.2(0.5) 0.4(0.4) 0.27

a

a

a

== + + == + + == + + =

Each entry represents the probability that a grandchild has a certain income level given his or her grandparents’ income level.

89. a b

Ac d

=

If 0D ad bc= − ≠ , then 0a ≠ and 0d ≠ , or 0b ≠ and 0c ≠ . Assuming the former, then

( )

( )

11

1 1

1

2 1 2

1

1 0

0 1

1 0

0 1

1 0

0 1

1 0

0 1

ba a

a

ba a

bc ca a

ba a

ad bc ca a

a b

c d

R rc d

R c r rd

−

→ = ⋅

→ = − ⋅ +− −

→−

( )

( )

1

2 2

1( )

1 2 1

1 0

0 1

1 0

0 1

1 0

0 1

1 0

0 1

ba a a

ad bcc aad bc ad bc

bc ba a ad bc ad bc b

ac aad bc ad bc

d bad bc ad bc

c aad bc ad bc

d bD D

c aD D

R r

R r r

−−− −

−− −

−− −

−− −

−− −

→ = ⋅

+→ = − ⋅ +

→

−→

−

Thus, 1 1d bD D

c aD D

d bA

c aD− − −

= =− −

where D ad bc= − .

90. Answers will vary.

91. Since the product is found by multiplying the components from the columns of the first matrix


1096


by the components in the rows of the second matrix and then adding those products, then the number of columns in the first must equal the number of rows in the second.

92. For real numbers this you multiply both sides by

the multiplicative inverse of c, 1c , c not equal 0.

Since c times 1c results in 1, the multiplicative

identity, then a = b. For matrices, this would hold true as long as C has an inverse. Then you

would multiply both sides by 1C− to get:

1 1

AC BC

ACC BCC

AI BI

A B

− −

==

==

93. If the inverse of A exists:

1 1

0

0

0

0

AX

A AX A

IX

X

− −

==

==

If the inverse of A does not exist, then A is singular and you would not be able to multiply A by its inverse to give the identity inverse and X would have no solution.

Section 11.5

1. True

2. True

3. ( )( )

4 3 2 2 2

22

3 6 3 3 2 1

3 1

x x x x x x

x x

+ + = + +

= +

4. True

5. The rational expression 2 1

x

x − is proper, since

the degree of the numerator is less than the degree of the denominator.

6. The rational expression 3

5 2

1

x

x

+−

is proper, since

the degree of the numerator is less than the degree of the denominator.


2

5

4

x

x

+−

is improper, so

perform the division:

2 2

2

1

4 5

4

9

x x

x

− +

−

The proper rational expression is: 2

2 2

5 91

4 4

x

x x

+ = +− −


2

3 2

1

x

x

−−

is improper, so


2 2

2

3

1 3 2

3 3

1

x x

x

− −

−


2 2

3 2 13

1 1

x

x x

− = +− −


25 2 1

4

x x

x

+ −−

is improper,

so perform the division:

2 3

3

5

4 5 2 1

5 20

22 1

x

x x x

x x

x

− + −−

−


2 25 2 1 22 1

4 45x x x

x xx+ − −

− −= +


33 2

8

x x

x

+ −+

is improper,

so perform the division:

3 4 2

4

2

3

8 3 2

3 24

24 2

x

x x x

x x

x x

+ + −+

− −

The proper rational expression is: 4 2 2

3 33 2 24 2

8 83x x x x

x xx+ − − −

+ += +

Section 11.5: Partial Fraction Decomposition

1097



2

( 1)( 4)( 3) 12

x x x xx x x x

− −=+ − + −

is improper, so


2 2

2

1

12 0

122 12

x x x x

x xx

+ − − ++ −

− +

The proper rational expression is:

2( 1) 2 12 2( 6)

( 4)( 3) ( 4)( 3)121 1x x x x

x x x xx x

− − + − −+ − + −+ −

= + = +


2 22 ( 4) 2 8

1 1

x x x x

x x

+ +=+ +

is

improper, so perform the division:

2 3

3

2

1 2 8

2 26

x

x x x

x xx

+ ++

The proper rational expression is:

2

2 22 ( 4) 6

1 12x x x

x xx+

+ += +

13. Find the partial fraction decomposition: 4

( 1) 1

4( 1) 1

( 1) ( 1)

4 ( 1)

A Bx x x x

A Bx x x x

x x x x

A x Bx

= +− −

+− −

− = −

= − +

Let 1x = , then 4 (0)4A B

B= +=

Let 0x = , then 4 ( 1) (0)4

A BA

= − += −

4 4 4( 1) 1x x x x

−= +− −

14. Find the partial fraction decomposition 3

( 2)( 1) 2 1x A B

x x x x= +

+ − + −

Multiplying both sides by ( 2)( 1)x x+ − , we

obtain: 3 ( 1) ( 2)x A x B x= − + +

Let 1x = , then 3(1) (0) (3)3 3

1

A BBB

= +==

Let 2x = − , then 3(– 2) ( 3) (0)3 6

2

A BAA

= − +− = −

=

3 2 1( 2)( 1) 2 1

xx x x x

= ++ − + −

15. Find the partial fraction decomposition:

2 2

2 22 2

2

1

( 1) 1

1

( 1) 1( 1) ( 1)

1 ( 1) ( )

A Bx Cxx x x

A Bx Cxx x x

x x x x

A x Bx C x

+= ++ +

+++ +

+ = +

= + + +

Let 0x = , then 21 (0 1) ( (0) )(0)

1

A B C

A

= + + +=

Let 1x = , then 21 (1 1) ( (1) )(1)

1 2

1 2(1)

1

A B C

A B C

B C

B C

= + + += + += + +

+ = −

Let 1x = − , then 21 (( 1) 1) ( ( 1) )( 1)

1 (1 1) ( )( 1)1 21 2(1)

1

A B CA B C

A B CB C

B C

= − + + − + −= + + − + −= + −= + −

− = −

Solve the system of equations: 11

2 21

B CB C

BB

+ = −− = −

= −= −

1 10

CC

− + = −=

2 2

1 1

( 1) 1

xxx x x

−= ++ +


2 21

1( 1)( 4) 4

A Bx Cxx x x

+= +++ + +

Multiplying both sides by 2( 1)( 4)x x+ + , we

obtain: 21 ( 4) ( )( 1)A x Bx C x= + + + +


1098


Let 1x = − , then 1 (5) ( ( 1) )(0)

5 1

15

A B C

A

A

= + − +=

=

Let 1x = , then

( )

21 (1 4) ( (1) )(1 1)

1 5 ( )(2)

1 5 1/ 5 2 2

1 1 2 2

0 2 2

0

A B C

A B C

B C

B C

B C

B C

= + + + += + += + += + += += +

Let 0x = , then

( )

21 (0 4) ( (0) )(0 1)

1 4

1 4 1/ 5

41

515

A B C

A C

C

C

C

= + + + += += +

= +

=

Since 0B C+ = , we have that 1

05

15

B

B

+ =

= −

1 1 15 5 5

2 2

11( 1)( 4) 4

x

xx x x

− += +

++ + +


( 1)( 2) 1 2x A B

x x x x= +

− − − −

Multiplying both sides by ( 1)( 2)x x− − , we

obtain: ( 2) ( 1)x A x B x= − + −

Let 1x = , then 1 (1 2) (1 1)

1

1

A B

A

A

= − + −= −= −

Let 2x = , then 2 (2 2) (2 1)

2

A B

B

= − + −=

1 2

( 1)( 2) 1 2x

x x x x−= +

− − − −


( 2)( 4) 2 4x A B

x x x x= +

+ − + −


obtain: 3 ( 4) ( 2)x A x B x= − + +

Let 2x = − , then 3( 2) ( 2 4) ( 2 2)

6 6

1

A B

A

A

− = − − + − +− = −

=

Let 4x = , then 3(4) (4 4) (4 2)

12 6

2

A B

B

B

= − + +==

3 1 2( 2)( 4) 2 4

xx x x x

= ++ − + −


2 21 1( 1) ( 1) ( 1)

x A B Cx xx x x

= + +− +− + −

Multiplying both sides by 2( 1) ( 1)x x− + , we

obtain: 2 2( 1)( 1) ( 1) ( 1)x A x x B x C x= − + + + + −

Let 1x = , then 2 2

2

1 (1 1)(1 1) (1 1) (1 1)

1 (0)(2) (2) (0)

1 2

12

A B C

A B C

B

B

= − + + + + −

= + +=

=

Let 1x = − , then 2 2

2

( 1) ( 1 1)( 1 1) ( 1 1) ( 1 1)

1 ( 2)(0) (0) ( 2)

1 4

14

A B C

A B C

C

C

− = − − − + + − + + − −

= − + + −=

=

Let 0x = , then 2 20 (0 1)(0 1) (0 1) (0 1)

0

1 1 32 4 4

A B C

A B C

A B C

A

= − + + + + −= − + += +

= + =

3 1 124 2 4

2 21 1( 1) ( 1) ( 1)

xx xx x x

= + +− +− + −


2 21

2( 2)

x A B Cx xx x x

+ = + +−−

Multiplying both sides by 2( 2)x x − , we obtain: 21 ( 2) ( 2)x Ax x B x Cx+ = − + − +


1099


Let 0x = , then 20 1 (0)(0 2) (0 2) (0)

1 2

12

A B C

B

B

+ = − + − += −

= −

Let 2x = , then 22 1 (2)(2 2) (2 2) (2)

3 4

34

A B C

C

C

+ = − + − +=

=

Let 1x = , then 2

2

1 3 32

2 4 4

A B C

A B C

A

= − − += − + −

= − − + − = −

3 314 2 4

2 2

12( 2)

xx xx x x

− −+ = + +−−


3 2

2 2

1 1

8 ( 2)( 2 4)1

2( 2)( 2 4) 2 4

x x x x

A Bx Cxx x x x x

=− − + +

+= +−− + + + +

Multiplying both sides by 2( 2)( 2 4)x x x− + + ,

we obtain: 21 ( 2 4) ( )( 2)A x x Bx C x= + + + + −

Let 2x = , then

( )21 2 2(2) 4 ( (2) )(2 2)

1 12

112

A B C

A

A

= + + + + −

=

=

Let 0x = , then

( )

( )

2

23

1 0 2(0) 4 ( (0) )(0 2)

1 4 2

1 4 1/12 2

2

13

A B C

A C

C

C

C

= + + + + −

= −= −

− =

= −

Let 1x = , then

( )

( )

21 1 2(1) 4 ( (1) )(1 2)

1 7

11 7 1/12

31

12

A B C

A B C

B

B

= + + + + −

= − −

= − +

= −

( )

1 1112 312

3 2

1 112 12

2

128 2 4

4

2 2 4

x

xx x xx

x x x

− −= +

−− + +− +

= +− + +


3 2 2

2 4 2 411 ( 1)( 1) 1

x x A Bx Cxx x x x x x

+ + += = +−− − + + + +

Multiplying both sides by 2( 1)( 1)x x x− + + , we

obtain: 22 4 ( 1) ( )( 1)x A x x Bx C x+ = + + + + −

Let 1x = , then

( ) ( )22(1) 4 1 1 1 (1) (1 1)

6 3

2

A B C

A

A

+ = + + + + −

==

Let 0x = , then

( )22(0) 4 0 0 1 ( (0) )(0 1)

4

4 2

2

A B C

A C

C

C

+ = + + + + −

= −= −= −

Let 1x = − , then

( )22( 1) 4 ( 1) ( 1) 1 ( ( 1) )( 1 1)

2 2 2

2 2 2 2( 2)

2 4

2

A B C

A B C

B

B

B

− + = − + − + + − + − −

= + −= + − −= −= −

3 22 4 2 2 2

11 1

x xxx x x

+ − −= +−− + +


2 2 2 21 1( 1) ( 1) ( 1) ( 1)

x A B C Dx xx x x x

= + + +− +− + − +

Multiplying both sides by 2 2( 1) ( 1)x x− + , we obtain:

2 2 2

2 2

( 1)( 1) ( 1)

( 1) ( 1) ( 1)

x A x x B x

C x x D x

= − + + +

+ − + + −


1100


Let 1x = , then 2 2 2

2 2

1 (1 1)(1 1) (1 1)

(1 1) (1 1) (1 1)

1 4

14

A B

C D

B

B

= − + + +

+ − + + −=

=

Let 1x = − , then 2 2 2

2 2

( 1) ( 1 1)( 1 1) ( 1 1)

( 1 1) ( 1 1) ( 1 1)

1 4

14

A B

C D

D

D

− = − − − + + − +

+ − − − + + − −=

=

Let 0x = , then 2 2 2

2 2

0 (0 1)(0 1) (0 1)

(0 1) (0 1) (0 1)

0

1 1 14 4 2

A B

C D

A B C D

A C B D

A C

= − + + +

+ − + + −= − + + +

− = +

− = + =

Let 2x = , then 2 2 2

2 2

2 (2 1)(2 1) (2 1)

(2 1) (2 1) (2 1)

4 9 9 3

9 3 4 9

1 1 39 3 4 9

4 4 21

32

A B

C D

A B C D

A C B D

A C

A C

= − + + +

+ − + + −= + + +

+ = − −

+ = − − =

+ =

Solve the system of equations: 1213 2

A C

A C

− =

+ =

4 114

3 14 2

14

A

A

C

C

=

=

+ =

= −

1 1 1 124 4 4 4

2 2 2 21 1( 1) ( 1) ( 1) ( 1)

xx xx x x x

−= + + +

− +− + − +


2 2 2 21

2( 2) ( 2)

x A B C Dx xx x x x

+ = + + +−− −

Multiplying both sides by 2 2( 2)x x − , we obtain: 2 2 2 21 ( 2) ( 2) ( 2)x Ax x B x Cx x Dx+ = − + − + − +

Let 0x = , then 2 2

2 2

0+1 (0)(0 2) (0 2)

(0) (0 2) (0)

1 4

14

A B

C D

B

B

= − + −

+ − +=

=

Let 2x = , then 2 2

2 2

2 1 (2)(2 2) (2 2)

(2) (2 2) (2)

3 4

34

A B

C D

D

D

+ = − + −

+ − +=

=

Let 1x = , then 2 2

2 2

1 1 (1)(1 2) (1 2)

(1) (1 2) (1)2

21 3

2 14 4

A B

C DA B C D

A C B D

A C

+ = − + −+ − +

= + − +− = − −

− = − − =

Let 3x = , then 2 2

2 2

3 1 (3)(3 2) (3 2)

(3) (3 2) (3)4 3 9 9

3 9 4 9

1 33 9 4 9 3

4 43 1

A B

C DA B C D

A C B D

A C

A C

+ = − + −+ − +

= + + ++ = − −

+ = − − = −

+ = −

Solve the system of equations: 1

3 1

A C

A C

− =+ = −

( )

1

1

3 1

1 3 1

A C

A C

A C

C C

− == +

+ = −+ + = −

4 2

12

C

C

= −

= −


1101


1 11 1

2 2A C= + = − + =

31 1 12 4 2 4

2 2 2 2

12( 2) ( 2)

xx xx x x x

−+ = + + +−− −


2 23

2 1( 2)( 1) ( 1)

x A B Cx xx x x

− = + ++ ++ + +

Multiplying both sides by 2( 2)( 1)x x+ + , we

obtain: 23 ( 1) ( 2)( 1) ( 2)x A x B x x C x− = + + + + + +

Let 2x = − , then 22 3 ( 2 1) ( 2 2)( 2 1) ( 2 2)

55

A B CA

A

− − = − + + − + − + + − +− =

= −

Let 1x = − , then 21 3 ( 1 1) ( 1 2)( 1 1) ( 1 2)

44

A B CC

C

− − = − + + − + − + + − +− =

= −

Let 0x = , then 20 3 (0 1) (0 2)(0 1) (0 2)

3 2 2

3 5 2 2( 4)

2 10

5

A B C

A B C

B

B

B

− = + + + + + +− = + +− = − + + −

==

2 2

3 5 5 42 1( 2)( 1) ( 1)

xx xx x x

− − −= + ++ ++ + +


2 22 1( 2)( 1) ( 1)

x x A B Cx xx x x

+ = + ++ −+ − −

Multiplying both sides by 2( 2)( 1)x x+ − , we obtain:

2 2( 1) ( 2)( 1) ( 2)x x A x B x x C x+ = − + + − + +

Let 2x = − , then 2 2( 2) ( 2) ( 2 1) ( 2 2)( 2 1)

( 2 2)

2 9

29

A B

C

A

A

− + − = − − + − + − −+ − +

=

=

Let 1x = , then 2 21 1 (1 1) (1 2)(1 1) (1 2)

2 3

23

A B C

C

C

+ = − + + − + +=

=

Let 0x = , then 2 20 0 (0 1) (0 2)(0 1) (0 2)

0 2 2

2 2

2 2 142 2

9 3 979

A B C

A B C

B A C

B

B

+ = − + + − + += − += +

= + =

=

72 229 9 3

2 22 1( 2)( 1) ( 1)

x xx xx x x

+ = + ++ −+ − −


2 2 2 24

( 4) 4

x A B Cx Dxx x x x

+ += + ++ +

Multiplying both sides by 2 2( 4)x x + , we obtain: 2 2 24 ( 4) ( 4) ( )x Ax x B x Cx D x+ = + + + + +

Let 0x = , then

( )2 2 20 4 (0)(0 4) (0 4) (0) (0)4 4

1

A B C DB

B

+ = + + + + +==

Let 1x = , then 2 2 21 4 (1)(1 4) (1 4) ( (1) )(1)

5 5 55 5 5

A B C DA B C DA C D

+ = + + + + += + + += + + +

5 0A C D+ + =

Let 1x = − , then 2 2

2

1 4 ( 1)(( 1) 4) (( 1) 4)

( ( 1) )( 1)3 5 53 5 5

A B

C DA B C DA C D

− + = − − + + − ++ − + −

= − + − += − + − +

5 2A C D− − + = −

Let 2x = , then 2 2 22 4 (2)(2 4) (2 4) ( (2) )(2)

6 16 8 8 46 16 8 8 4

A B C DA B C DA C D

+ = + + + + += + + += + + +

16 8 4 2A C D+ + = −


1102


Solve the system of equations: 5 05 2

2 21

A C DA C D

DD

+ + =− − + = −

= −= −

5 1 01 5

A CC A

+ − == −

16 8(1 5 ) 4( 1) 216 8 40 4 2

24 614

A AA A

A

A

+ − + − = −+ − − = −

− = −

=

1 5 1

1 5 14 4 4

C = − = − = −

( )

1 14 4

2 2 2 2

1 14 4

2 2

14 1

( 4) 4

41

4

xxxx x x x

x

x x x

− −+ = + ++ +

− += + +

+


2 2 2 2

10 21( 1) ( 2) ( 1) 2

x x A B Cx Dxx x x x

+ += + +−− + − +

Multiply both sides by 2 2( 1) ( 2)x x− + : 2 2 2

2

10 2 ( 1)( 2) ( 2)

( )( 1)

x x A x x B x

Cx D x

+ = − + + +

+ + −

Let 1x = , then

( )2 2 2

2

10(1) 2(1) (1 1)(1 2) (1 2)

(1) (1 1)

12 3

4

A B

C D

B

B

+ = − + + +

+ + −==

Let 0x = , then

( )2 2 2

2

10(0) 2(0) (0 1)(0 2) (0 2)

(0) (0 1)

0 2 2

0 2 8

2 8

2 8

A B

C D

A B D

A D

A D

D A

+ = − + + +

+ + −= − + += − + +

− == −

Let 1x = − , then 2 2

2

2

10( 1) 2( 1) ( 1 1)(( 1) 2)

(( 1) 2)

( ( 1) )( 1 1)

8 6 3 4 4

8 6 12 4 4

6 4 4 4

A

B

C D

A B C D

A C D

A C D

− + − = − − − +

+ − +

+ − + − −= − + − += − + − +

− − + = −

Let 2x = , then 2 2 2

2

10(2) 2(2) (2 1)(2 2) (2 2)

( (2) )(2 1)

44 6 6 2

44 6 24 2

6 2 20

A B

C D

A B C D

A C D

A C D

+ = − + + +

+ + −= + + += + + +

+ + =

Solve the system of equations (Substitute for D): 2 8

6 4 4 4

6 4 4(2 8) 4

2 4 28

2 14

D A

A C D

A C A

A C

A C

= −− − + = −

− − + − = −− =− =

( )6 2 20

6 2 2 8 20

8 2 28

A C D

A C A

A C

+ + =+ + − =

+ =

Add the equations and solve:

2 14

8 2 28

A C

A C

− =+ =

9 42

143

A

A

=

=

14 282 14 14

3 3143

C A

C

= − = − = −

= −

14 42 8 2 8

3 3D A= − = − =

14 14 423 3 3

2 2 2 2

10 2 41( 1) ( 2) ( 1) 2

xx xxx x x x

− ++ = + +−− + − +


1103



2 2

2 31( 1)( 2 4) 2 4

x x A Bx Cxx x x x x

+ + += +++ + + + +

Multiplying both sides by 2( 1)( 2 4)x x x+ + + , we obtain:

2 22 3 ( 2 4) ( )( 1)x x A x x Bx C x+ + = + + + + +

Let 1x = − , then 2 2( 1) 2( 1) 3 (( 1) 2( 1) 4)

( ( 1) )( 1 1)

2 3

23

A

B C

A

A

− + − + = − + − ++ − + − +

=

=

Let 0x = , then

( )

2 20 2(0) 3 (0 2(0) 4) ( (0) )(0 1)

3 4

3 4 2/3

13

A B C

A C

C

C

+ + = + + + + += += +

=

Let 1x = , then

( ) ( )

2 2

14 2 23 3 3

13

1 2(1) 3 (1 2(1) 4) ( (1) )(1 1)

6 7 2 2

6 7 2 / 3 2 2 1/ 3

2 6

A B C

A B C

B

B

B

+ + = + + + + += + += + +

= − − =

=

2 1 123 3 3

2 2

2 13 3

2

2 31( 1)( 2 4) 2 4

( 1)

1 2 4

xx xxx x x x x

x

x x x

++ + = +++ + + + +

+= +

+ + +


2 2

11 18

( 3 3) 3 3

x x A Bx Cxx x x x x

− − += ++ + + +

Multiplying both sides by 2( 3 3)x x x+ + 2( 1)( 2 4)x x x+ + + , we obtain:

2 211 18 ( 3 3) ( )x x A x x Bx C x− − = + + + +

Let 0x = , then

( ) ( )2 20 11(0) 18 0 3(0) 3 (0) (0)

18 3

6

A B C

A

A

− − = + + + +

− == −

Let 1x = , then

( ) ( )2 21 11(1) 18 1 3(1) 3 (1) (1)

28 7

28 7( 6)

14

A B C

A B C

B C

B C

− − = + + + +

− = + +− = − + ++ =

Let 1x = − , then

( )( )

2 2( 1) 11( 1) 18 ( 1) 3( 1) 3

( 1) ( 1)

6

6 6

0

A

B C

A B C

B C

B C

− − − − = − + − +

+ − + −− = + −− = − + −

− =

Add the last two equations and solve: 14

0

B C

B C

+ =− =

2 14

7

B

B

==

14

7 14

7

B C

C

C

+ =+ =

=

2

2 2

11 18 6 7 7

( 3 3) 3 3

x x xxx x x x x

− − − += ++ + + +


(3 2)(2 1) 3 2 2 1x A B

x x x x= +

− + − +

Multiplying both sides by (3 2)(2 1)x x− + , we

obtain: (2 1) (3 2)x A x B x= + + −

Let 12x = − , then

( )( ) ( )( )12 1/ 2 1 3 1/ 2 2

21 72 2

17

A B

B

B

− = − + + − −

− = −

=

Let 23

x = , then

( )( ) ( )( )22 2 / 3 1 3 2 / 3 2

32 73 3

27

A B

A

A

= + + −

=

=

2 17 7

(3 2)(2 1) 3 2 2 1x

x x x x= +

− + − +


1104



(2 3)(4 1) 2 3 4 1A B

x x x x= +

+ − + −

Multiplying both sides by (2 3)(4 1)x x+ − , we

obtain: 1 (4 1) (2 3)A x B x= − + +

Let 32

x = − , then

3 31 4 1 2 3

2 2

1 7

17

A B

A

A

= − − + − +

= −

= −

Let 14

x = , then

1 11 4 1 2 3

4 4

71

227

A B

B

B

= − + +

=

=

1 27 71

(2 3)(4 1) 2 3 4 1x x x x

−= +

+ − + −


2 ( 3)( 1) 3 12 3

x x A Bx x x xx x

= = ++ − + −+ −


obtain: ( 1) ( 3)x A x B x= − + +

Let 1x = , then 1 (1 1) (1 3)1 4

14

A BB

B

= − + +=

=

Let 3x = − , then 3 ( 3 1) ( 3 3)3 4

34

A BA

A

− = − − + − +− = −

=

3 14 4

2 3 12 3

xx xx x

= ++ −+ −

34. Find the partial fraction decomposition: 2 2

28 8

( 1)( 2)( 3)( 1)( 5 6)

1 2 3

x x x xx x xx x xA B C

x x x

− − − −=+ + ++ + +

= + ++ + +

Multiplying both sides by ( 1)( 2)( 3)x x x+ + + , we obtain:

2 8 ( 2)( 3) ( 1)( 3)

( 1)( 2)

x x A x x B x x

C x x

− − = + + + + ++ + +

Let 1x = − , then 2( 1) ( 1) 8 ( 1 2)( 1 3)

( 1 1)( 1 3)

( 1 1)( 1 2)

6 2

3

A

B

C

A

A

− − − − = − + − ++ − + − +

+ − + − +− =

= −

Let 2x = − , then 2( 2) ( 2) 8 ( 2 2)( 2 3)

( 2 1)( 2 3) ( 2 1)( 2 2)

22

AB

CB

B

− − − − = − + − ++ − + − +

+ − + − +− = −

=

Let 3x = − , then 2( 3) ( 3) 8 ( 3 2)( 3 3)

( 3 1)( 3 3) ( 3 1)( 3 2)

4 22

AB

CC

C

− − − − = − + − ++ − + − +

+ − + − +==

2

28 3 2 2

1 2 3( 1)( 5 6)

x xx x xx x x

− − −= + ++ + ++ + +


2 2 2 2 2

2 3

( 4) 4 ( 4)

x x Ax B Cx D

x x x

+ + + += ++ + +

Multiplying both sides by 2 2( 4)x + , we obtain: 2 2

2 3 2

2 3 2

2 3 ( )( 4)

2 3 4 4

2 3 (4 ) 4

x x Ax B x Cx D

x x Ax Bx Ax B Cx D

x x Ax Bx A C x B D

+ + = + + + +

+ + = + + + + +

+ + = + + + + +

0A = ; 1B = ;

4 2

4(0) 2

2

A C

C

C

+ =+ =

=

4 3

4(1) 3

1

B D

D

D

+ =+ =

= −

2

2 2 2 2 2

2 3 1 2 1

( 4) 4 ( 4)

x x x

x x x

+ + −= ++ + +


1105



2 2 2 2 2

1

( 16) 16 ( 16)

x Ax B Cx D

x x x

+ + += ++ + +

Multiplying both sides by 2 2( 16)x + , we obtain: 3 2

3 3 2

3 3 2

1 ( )( 16)

1 16 16

1 (16 ) 16

x Ax B x Cx D

x Ax Bx Ax B Cx D

x Ax Bx A C x B D

+ = + + + +

+ = + + + + +

+ = + + + + +

1A = ; 0B = ;

16 0

16(1) 0

16

A C

C

C

+ =+ =

= −

16 1

16(0) 1

1

B D

D

D

+ =+ =

=

3

2 2 2 2 2

1 16 1

( 16) 16 ( 16)

x x x

x x x

+ − += ++ + +


3 2

7 3 7 3( 3)( 1)2 3

3 1

x xx x xx x xA B Cx x x

+ +=− +− −

= + +− +

Multiplying both sides by ( 3)( 1)x x x− + , we obtain: 7 3 ( 3)( 1) ( 1) ( 3)x A x x Bx x Cx x+ = − + + + + −

Let 0x = , then 7(0) 3 (0 3)(0 1) (0)(0 1) (0)(0 3)

3 3

1

A B C

A

A

+ = − + + + + −= −= −

Let 3x = , then 7(3) 3 (3 3)(3 1) (3)(3 1) (3)(3 3)

24 12

2

A B C

B

B

+ = − + + + + −==

Let 1x = − , then 7( 1) 3 ( 1 3)( 1 1) ( 1)( 1 1)

( 1)( 1 3)

4 4

1

A B

C

C

C

− + = − − − + + − − ++ − − −

− == −

3 2

7 3 1 2 13 12 3

xx x xx x x

+ − −= + +− +− −


5 4 4

2 3 4

1 1

( 1)

1

x x

x x x x

A B C D Ex xx x x

+ +=− −

= + + + +−

Multiplying both sides by 4 ( 1)x x − , we obtain: 3 3 2

4

1 ( 1) ( 1) ( 1)

( 1)

x Ax x Bx x Cx x

D x Ex

+ = − + − + −

+ − +

Let 0x = , then 4 3 2

4

0 1 0 (0 1) 0 (0 1) 0(0 1)

(0 1) 0

A B C

D E

+ = ⋅ − + ⋅ − + ⋅ −

+ − + ⋅

1

1

D

D

= −= −

Let 1x = , then 4 3 2

4

1 1 1 (1 1) 1 (1 1) 1(1 1)

(1 1) 1

A B C

D E

+ = ⋅ − + ⋅ − + ⋅ −

+ − + ⋅

2 E=

Let 1x = − , then 4 3 2

4

( 1) 1 ( 1) ( 1 1) ( 1) ( 1 1)

( 1)( 1 1) ( 1 1) ( 1)

A B

C D E

− + = − − − + − − −

+ − − − + − − + −

0 2 2 2 2

0 2 2 2 2( 1) 2

2 2 2 4

2

A B C D E

A B C

A B C

A B C

= − + − += − + − − +

− + = −− + = −

Let 2x = , then 4 3 2

4

2 1 2 (2 1) 2 (2 1) 2(2 1)

(2 1) 2

A B C

D E

+ = ⋅ − + ⋅ − + ⋅ −

+ − + ⋅

9 8 4 2 16

9 8 4 2 ( 1) 16(2)

8 4 2 22

4 2 11

A B C D E

A B C

A B C

A B C

= + + + += + + + − +

+ + = −+ + = −

Let 2x = − , then 4 3 2

4

( 22) 1 ( 2) ( 2 1) ( 2) ( 2 1)

( 2)( 2 1) ( 2 1) ( 2)

A B

C D E

− + = − − − + − − −

+ − − − + − − + −

7 24 12 6 3 16

7 24 12 6 3( 1) 81(2)

42 24 12 6

14 8 4 2

A B C D E

A B C

A B C

A B C

− = − + − +− = − + − − +

− = − +− = − +


1106


Solve the system of equations by using a matrix equation:

2

4 2 11

8 4 2 14

A B C

A B C

A B C

− + = −+ + = −

− + = −

1

1 1 1 2

4 2 1 11

8 4 2 14

1 1 1 2 2

4 2 1 11 1

8 4 2 14 1

A

B

C

A

B

C

−

− −= −

− −

− − −= − = −

− − −

So, 2A = − , 1B = − , and 1C = − . Thus,

3 3

5 4 4

2 3 4

1 1

( 1)

2 1 1 1 21

x x

x x x x

x xx x x

+ +=− −

− − − −= + + + +−

39. Perform synthetic division to find a factor:

2 1 4 5 2

2 4 2

1 2 1 0

− −−

−

3 2 2

2

4 5 2 ( 2)( 2 1)

( 2)( 1)

x x x x x x

x x

− + − = − − +

= − −

Find the partial fraction decomposition:

2 2

3 2 2

2

4 5 2 ( 2)( 1)

2 1 ( 1)

x x

x x x x x

A B Cx x x

=− + − − −

= + +− − −

Multiplying both sides by 2( 2)( 1)x x− − , we obtain:

2 2( 1) ( 2)( 1) ( 2)x A x B x x C x= − + − − + − Let 2x = , then

2 22 (2 1) (2 2)(2 1) (2 2)

4

A B C

A

= − + − − + −=

Let 1x = , then

2 21 (1 1) (1 2)(1 1) (1 2)

1

1

A B C

C

C

= − + − − + −= −= −

Let 0x = , then 2 20 (0 1) (0 2)(0 1) (0 2)

0 2 2

0 4 2 2( 1)

2 6

3

A B C

A B C

B

B

B

= − + − − + −= + −= + − −

− == −

2

3 2 2

4 3 12 14 5 2 ( 1)

xx xx x x x

− −= + +− −− + − −

40. Perform synthetic division to find a factor:

1 1 1 5 3

1 2 3

1 2 3 0

−−

−

3 2 2

2

5 3 ( 1)( 2 3)

( 3)( 1)

x x x x x x

x x

+ − + = − + −

= + −


2 2

3 2 2

2

1 1

5 3 ( 3)( 1)

3 1 ( 1)

x x

x x x x x

A B Cx x x

+ +=+ − + + −

= + ++ − −

Multiplying both sides by 2( 3)( 1)x x+ − , we obtain:

2 21 ( 1) ( 3)( 1) ( 3)x A x B x x C x+ = − + + − + +

Let 3x = − , then 2 2( 3) 1 ( 3 1) ( 3 3)( 3 1)

( 3 3)

10 16

58

A B

C

A

A

− + = − − + − + − −+ − +

=

=

Let 1x = , then 2 21 1 (1 1) (1 3)(1 1) (1 3)

2 4

12

A B C

C

C

+ = − + + − + +=

=


1107


Let 0x = , then 2 20 1 (0 1) (0 3)(0 1) (0 3)

1 3 3

5 11 3 3

8 29

3838

A B C

A B C

B

B

B

+ = − + + − + += − +

= − +

=

=

5 3 128 8 2

3 2 2

13 15 3 ( 1)

xx xx x x x

+ = + ++ −+ − + −


2 3 2 2 2 2 3( 16) 16 ( 16) ( 16)

x Ax B Cx D Ex F

x x x x

+ + += + ++ + + +

Multiplying both sides by 2 3( 16)x + , we obtain: 3 2 2 2

3 4 2 3 2

3 5 4 3 2

( )( 16) ( )( 16)

( )( 32 256) 16 16

32 32 256

x Ax B x Cx D xEx F

x Ax B x x Cx DxCx D Ex F

x Ax Bx Ax Bx Ax

= + + + + ++ +

= + + + + ++ + + +

= + + + +3 2 256

16 16B Cx DxCx D Ex F

+ + ++ + + +

3 5 4 3 2(32 ) (32 )

(256 16 )

(256 16 )

x Ax Bx A C x B D x

A C E x

B D F

= + + + + ++ + +

+ + +

0; 0A B= = ; 32 1

32(0) 1

1

A C

C

C

+ =+ =

=

32 0

32(0) 0

0

B D

D

D

+ =+ =

=

256 16 0

256(0) 16(1) 0

16

A C E

E

E

+ + =+ + =

= −

256 16 0

256(0) 16(0) 0

0

B D F

F

F

+ + =+ + =

=

3

2 3 2 2 2 3

16

( 16) ( 16) ( 16)

x x x

x x x

−= ++ + +


2 3 2 2 2 2 3( 4) 4 ( 4) ( 4)

x Ax B Cx D Ex F

x x x x

+ + += + ++ + + +

Multiplying both sides by 2 3( 4)x + , we obtain: 2 2 2 2

2 4 2 3 2

2 5 4 3 2

3 2

( )( 4) ( )( 4)

( )( 8 16)

4 4

8 8 16 16

4 4

x Ax B x Cx D x

Ex F

x Ax B x x Cx Dx

Cx D Ex F

x Ax Bx Ax Bx Ax B

Cx Dx Cx D Ex

= + + + + ++ +

= + + + + ++ + + +

= + + + + +

+ + + + +2 5 4 3 2(8 ) (8 )

(16 4 ) (16 4 )

F

x Ax Bx A C x B D x

A C E x B D F

+

= + + + + ++ + + + + +

0; 0A B= = ; 8 0

8(0) 0

0

A C

C

C

+ =+ =

=

8 1

8(0) 1

1

B D

D

D

+ =+ =

=

16 4 0

16(0) 4(0) 0

0

A C E

E

E

+ + =+ + =

=

16 4 0

16(0) 4(1) 0

4

B D F

F

F

+ + =+ + =

= −

2

2 3 2 2 2 3

1 4

( 4) ( 4) ( 4)

x

x x x

−= ++ + +


2

4 4( 3)(2 1) 3 2 12 5 3

A Bx x x xx x

= = +− + − +− −

Multiplying both sides by ( 3)(2 1)x x− + , we

obtain: 4 (2 1) ( 3)A x B x= + + −

Let 12

x = − , then

1 14 2 1 3

2 274 287

A B

B

B

= − + + − −

= −

= −


1108


Let 3x = , then 4 (2(3) 1) (3 3)

4 7

47

A B

A

A

= + + −=

=

847 7

2

43 2 12 5 3 x xx x

−= +

− +− −


2

4 4( 2)(2 1) 2 2 12 3 2

x x A Bx x x xx x

= = ++ − + −+ −

Multiplying both sides by ( 2)(2 1)x x+ − , we obtain: 4 (2 1) ( 2)x A x B x= − + +

Let 12

x = , then

1 1 14 2 1 2

2 2 2

52

345

A B

B

B

= − + +

=

=

Let 2x = − , then 4( 2) (2( 2) 1) ( 2 2)

8 5

85

A B

A

A

− = − − + − +− = −

=

8 45 5

2

42 2 12 3 2

xx xx x

= ++ −+ −


4 2 2

2

2 3 2 3

9 ( 3)( 3)

3 3

x x

x x x x x

A B C Dx x xx

+ +=− − +

= + + +− +

Multiplying both sides by 2 ( 3)( 3)x x x− + , we obtain:

2 2

2 3 ( 3)( 3) ( 3)( 3)

( 3) ( 3)

x Ax x x B x x

Cx x Dx x

+ = − + + − +

+ + + −

Let 0x = , then

2 2

2 0 3 0(0 3)(0 3) (0 3)(0 3)

0 (0 3) 0 (0 3)

3 9

13

A B

C D

B

B

⋅ + = ⋅ − + + − +

+ ⋅ + + ⋅ −= −

= −

Let 3x = , then

2 2

2 3 3 3(3 3)(3 3) (3 3)(3 3)

3 (3 3) 3 (3 3)

9 54

16

A B

C D

C

C

⋅ + = ⋅ − + + − +

+ ⋅ + + ⋅ −=

=

Let 3x = − , then

2

2

2( 3) 3 ( 3)( 3 3)( 3 3)

( 3 3)( 3 3)

( 3) ( 3 3)

( 3) ( 3 3)

3 54

118

A

B

C

D

D

D

− + = − − − − ++ − − − +

+ − − +

+ − − −− = −

=

Let 1x = , then

( ) ( ) ( )

2 2

2 1 3 1(1 3)(1 3) (1 3)(1 3)

1 (1 3) 1 (1 3)

5 8 8 4 2

5 8 8 1/ 3 4 1/ 6 2 1/18

8 2 15 8

3 3 916

89

29

A B

C D

A B C D

A

A

A

A

⋅ + = ⋅ − + + − +

+ ⋅ + + ⋅ −= − − + −= − − − + −

= − + + −

− =

= −

2 1 1 19 3 6 18

4 2 2

2 33 39

xx x xx x x

− −+ = + + +− +−


4 2 2

2

9 9

2 8 ( 2)( 2)( 2)

2 2 2

x x

x x x x x

A B Cx Dx x x

+ +=− − + − +

+= + +− + +

Multiplying both sides by 2( 2)( 2)( 2)x x x+ − + , we obtain:

2 2 29 ( 2)( 2) ( 2)( 2)

( )( 2)( 2)

x A x x B x x

Cx D x x

+ = + + + − ++ + − +


1109


Let 2x = , then 2 2 22 9 (2 2)(2 2) (2 2)(2 2)

( (2) )(2 2)(2 2)

13 24

1324

A B

C D

A

A

+ = + + + − ++ + − +

=

=

Let 2x = − , then 2 2

2

( 2) 9 (( 2) 2)( 2 2)

( 2 2)(( 2) 2)

( ( 2) )( 2 2)( 2 2)

13 24

1324

A

B

C D

B

B

− + = − + − +

+ − − − ++ − + − − − +

= −

= −

Let 0x = , then 2 2 20 9 (0 2)(0 2) (0 2)(0 2)

( (0) )(0 2)(0 2)

9 4 4 4

13 139 4 4 4

24 2414

4376

A B

C D

A B D

D

D

D

+ = + + + − ++ + − +

= − −

= − − −

= −

= −

Let 1x = , then 2 2 21 9 (1 2)(1 2) (1 2)(1 2)

( (1) )(1 2)(1 2)

10 9 3 3 3

39 13 710 3

8 8 23 0

0

A B

C D

A B C D

C

C

C

+ = + + + − ++ + − +

= − − −

= + − +

==

713 132624 24

4 2 2

92 22 8 2

xx xx x x

−−+ = + +− +− − +

47. 3 2

2

3

3 4

x x

x x

+ −+ −

Dividing:

( )

( )

2 3 2

3 2

2

2

2 3 4 0 3

3 4

2 4 3 2 6 8

10 11

xx x x x x

x x x

x xx x

x

−+ − + + −

− + −

− + −− − − +

−

3 2

2 2

3 10 112

3 4 3 4

x x xx

x x x x

+ − −= − ++ − + −

, 4,1x ≠ −


2

10 11 10 11( 4)( 1) 4 13 4

x x A Bx x x xx x

− −= = ++ − + −+ −

Multiplying both sides by ( 4)( 1)x x+ − , we obtain: 10 11 ( 1) ( 4)x A x B x− = − + + Let 1x = , then

( ) ( ) ( )10 1 11 1 1 1 4

1 5

15

A B

B

B

− = − + +− =

= −

Let 4x = − , then 10( 4) 11 (( 4) 1) ( 4 4)

51 5

515

A B

A

A

− − = − − + − +− = −

=

51 15 5

2

10 114 13 4

xx xx x

−− = ++ −+ −

Thus, 51 13 25 5

2

32

4 13 4

x xx

x xx x

−+ − = − + ++ −+ −

48. 3 2

2

3 1

5 6

x x

x x

− ++ +

Dividing:

( )

( )

2 3 2

3 2

2

2

8 5 6 3 0 1

5 6

8 6 1 8 40 48

34 49

xx x x x x

x x x

x xx x

x

−+ + − + +

− + +

− − +− − − −

+

3 2

2 2

3 1 34 498

5 6 5 6

x x xx

x x x x

− + += − ++ + + +

, 2, 3x ≠ − −


2

34 49 34 49( 2)( 3) 2 35 6

x x A Bx x x xx x

+ += = ++ + + ++ +

Multiplying both sides by ( 2)( 3)x x+ + , we obtain: 34 49 ( 3) ( 2)x A x B x+ = + + + Let 3x = − , then


1110


( ) ( ) ( )34 3 49 3 3 3 2

53

53

A B

B

B

− + = − + + − +− = −

=

Let 2x = − , then

( ) ( ) ( )34 2 49 2 3 2 2

19

A B

A

− + = − + + − +− =

2

34 49 53 192 35 6

xx xx x

+ −= ++ ++ +

Thus, 3 2

2

3 1 53 198

2 35 6

x xx

x xx x

− + −= − + ++ ++ +

49. 3

2 1

x

x +

Dividing:

( )2 3 2

3 2

0 1 0 0

0

xx x x x x

x x x

x

+ + + +

− + +−

3

2 21 1

x xx

x x

−= ++ +

Since 2 1x + is irreducible then we cannot go any further.

50. 3

2 4

x x

x

++

Dividing:

( )2 3 2

3 2

0 4 0

0 4

3

xx x x x x

x x x

x

+ + + +

− + +−

3

2 2

3

4 4

x x xx

x x

+ −= ++ +

Since 2 4x + is irreducible then we cannot go any further.

51. 4 3

2

5 4

4 4

x x x

x x

− + −+ +

Dividing:

( )

( )

2

2 4 3 2

4 3 2

3 2

3 2

2

4 7 4 4 0 5 4

4 4

4 9 4 16 16

7

x xx x x x x x

x x x

x x xx x x

x

− ++ + + − + −

− + +

− − +− − − −

( )2

17 4 7 28 28

11 32

xx x

x

+ −− + +

− −

4 32

2 2

5 4 11 324 7

4 4 4 4

x x x xx x

x x x x

− + − − −= − + ++ + + +

,

2x ≠ − Find the partial fraction decomposition:

2 2

11 32 11 32( 2)( 2) 24 4 ( 2)

x x A Bx x xx x x

− − − −= = ++ + ++ + +

Multiplying both sides by 2( 2)x + , we obtain:

11 32 ( 2)

11 32 2

11 32 (2 )

x A x B

x Ax A B

x Ax A B

− − = + +− − = + +− − = + +

Since the coefficient of x is A then 11A = − . Let 11A = − , then

32 2

32 2( 11)

10

A B

B

B

− = +− = − +− =

2 2

11 32 11 1024 4 ( 2)

xxx x x

− − − −= +++ + +

Thus, 4 3

22 2

5 4 11 104 7

24 4 ( 2)

x x xx x

xx x x

− + − − −= − + + +++ + +

52. 4 3

2

2

2 1

x x x

x x

+ − +− +

Dividing:

( )

( )

2

2 4 3 2

4 3 2

3 2

3 2

2

3 5 2 1 3 0 2

2

3 3 6 3

5

x xx x x x x x

x x x

x x xx x x

x

+ +− + + + − +

− − +

− −− − +

−( )2

4 2 5 10 5

6 3

xx x

x

+− − +

−

4 32

2 2

2 6 33 5

2 1 2 1

x x x xx x

x x x x

+ − + −= + + +− + − +

, 1x ≠


1111



2 2

6 3 6 3( 1)( 1) 12 1 ( 1)

x x A Bx x xx x x

− −= = +− − −− + −

Multiplying both sides by 2( 1)x − , we obtain:

6 3 ( 1)

6 3

6 3 ( )

x A x B

x Ax A B

x Ax A B

− = − +− = − +− = + − +

Since the coefficient of x is A then 6A = . Let 6A = , then

3

3 1(6)

3

A B

B

B

− = − +− = − +

=

2 2

6 3 6 312 1 ( 1)

xxx x x

− = +−− + −

Thus, 4 3

22 2

2 6 33 5

12 1 ( 1)

x x xx x

xx x x

+ − + = + + + +−− + −

53. 5 4 2

4 2

2

2 1

x x x

x x

+ − +− +

Dividing:

( )4 3 2 5 4 3 2

5 4 3 2

4 3 2

1 0 2 0 1 0 0 2

0 2 0

2 2

xx x x x x x x x x

x x x x x

x x x x

++ − + + + + − + +

− + − + +

+ − − +( )4 3 2

3 2

0 2 0 1

2 1

x x x x

x x x

− + − + +

− − +

5 4 2 3 2

4 2 4 2

2 2 11

2 1 2 1

x x x x x xx

x x x x

+ − + + − += + +− + − +

,

1, 1x ≠ − Find the partial fraction decomposition:

3 2

2 2 2 2

2 11 1( 1) ( 1) ( 1) ( 1)

x x x A B C Dx xx x x x

+ − + = + + ++ −+ − + −

Multiplying both sides by 2 2( 1) ( 1)x x+ − , we obtain:

3 2 2 2

2 2

2 1 ( 1)( 1) ( 1)

( 1) ( 1) ( 1)

x x x A x x B x

C x x D x

+ − + = + − + −

+ + − + +

Let 1x = − , then 3 2 2

2 2 2

2( 1) ( 1) ( 1) 1 ( 1 1)( 1 1)

( 1 1) ( 1 1) ( 1 1) ( 1 1)

A

B C D

− + − − − + = − + − −

+ − − + − + − − + − +

1 4

14

B

B

=

=

Let 1x = , then 3 2 2

2 2 2

2(1) (1) (1) 1 (1 1)(1 1)

(1 1) (1 1) (1 1) (1 1)

A

B C D

+ − + = + −

+ − + + − + +

3 4

34

D

D

=

=

Let 0x = , then 3 2 2

2 2 2

2(0) (0) (0) 1 (0 1)(0 1)

1 3 (0 1) (0 1) (0 1) (0 1)

4 41 3

14 4

0

A

C

A C

A C

+ − + = + −

+ − + + − + +

= + − +

= −

Let 2x = , then 3 2 2

2 2 2

2(2) (2) (2) 1 (2 1)(2 1)

1 3 (2 1) (2 1) (2 1) (2 1)

4 41 27

19 3 94 4

12 3 9

A

C

A C

A C

+ − + = + −

+ − + + − + +

= + + +

= +

0

12 3 9

12 3 9

12 12

1 and 1

A C

A C

A C

A A

A

A C

= −= +== +== =

3 2

4 2 2 2

1 32 1 1 14 4

1 12 1 ( 1) ( 1)

x x xx xx x x x

+ − + = + + ++ −− + + −

Thus, 5 4 2

4 2

2 2

21

2 11 3

14 4 1( 1) ( 1)

x x xx

x x

xx x

+ − + = +− +

+ + +−+ −


1112


54. 5 3 2

4 2

1

6 9

x x x

x x

− + ++ +

Dividing:

( )4 3 2 5 4 3 2

5 4 3 2

3 2

0 6 0 9 0 0 1

0 6 0 9

7 9 1

xx x x x x x x x x

x x x x x

x x x

+ + + + + − + + +

− + + + +

− + − +

5 3 2 3 2

4 2 4 2

1 7 9 1

6 9 6 9

x x x x x xx

x x x x

− + + − + − += ++ + + +

,

1, 1x ≠ − Find the partial fraction decomposition:

3 2

2 2 2 2 2

7 9 1

( 3)( 3) 3 ( 3)

x x x Ax B Cx D

x x x x

− + − + + += ++ + + +

Multiply

ing both sides by 2 2( 3)x + , we obtain: 3 2 2

3 2

3 2

7 9 1 ( )( 3) ( )

3 3

(3 ) (3 )

x x x Ax B x Cx D

Ax Bx Ax B Cx D

Ax Bx x A C x B D

− + − + = + + + +

= + + + + +

= + + + + +Then 7 and 1A B= − = .

9 3

1 3

9 3( 7) 12

1 3(1) 2

A C

B D

So

C C

and

D D

− = += +

− = − + =

= + = −

3 2

2 2 2 2 2

7 9 1 7 1 12 2

( 3)( 3) 3 ( 3)

x x x x x

x x x x

− + − + − + −= ++ + + +

Thus, 5 3 2

4 2 2 2 2

1 7 1 12 2

6 9 3 ( 3)

x x x x xx

x x x x

− + + − + −= + ++ + + +

Section 11.6

1. 3 2y x= + The graph is a line. x-intercept:

0 3 2

3 2

23

x

x

x

= += −

= −

y-intercept: ( )3 0 2 2y = + =

2

−2

y

x

(0, 2)

2 ,03

−

2

−2

2. 2 4y x= − The graph is a parabola. x-intercepts:

2

2

0 4

4

2, 2

x

x

x x

= −== − =

y-intercept: 20 4 4y = − = − The vertex has x-coordinate:

( )0

02 2 1b

xa

= − = − = .

The y-coordinate of the vertex is 20 4 4y = − = − .

Section 11.6: Systems of Nonlinear Equations

1113


3. 2 2

2 2

22

2 2

1

1

11 1

y x

x y

yx

= −− =

− =

The graph is a hyperbola with center (0, 0), transverse axis along the x-axis, and vertices at ( 1,0)− and (1,0) . The asymptotes are y x= − and y x= .

y

x−5(−1, 0)

5

−5

5(1, 0)

4. 2 2

2 2

22

22

2 2

4 4

4 44 4

14

12 1

x y

x y

xy

yx

+ =+ =

+ =

+ =

The graph is an ellipse with center (0,0) , major

axis along the x-axis, vertices at ( 2,0)− and

(2,0) . The graph also has y-intercepts at (0, 1)−

and (0,1) . y

x−5(−2, 0)

5

−5

5(2, 0)

(0, −1)

(0, 1)

5. 2 1

1

y x

y x

= += +

(0, 1) and (1, 2) are the intersection points.

Solve by substitution: 2

2

1 1

0

( 1) 0

0 or 1

1 2

x x

x x

x x

x x

y y

+ = +− =− =

= == =

Solutions: (0, 1) and (1, 2)

6. 2 1

4 1

y x

y x

= += +



2

1 4 1

4 0

( 4) 0

0 or 4

1 17

x x

x x

x x

x x

y y

+ = +− =− =

= == =



1114


7. 236

8

y x

y x

= −= −

(2.59, 5.41) and (5.41, 2.59) are the intersection points.


2 2

2

2

36 8

36 64 16

2 16 28 0

8 14 0

8 64 562

8 2 22

4 2

x x

x x x

x x

x x

x

− = −− = − +

− + =− + =

± −=

±=

= ±

( )( )

If 4 2, 8 4 2 4 2

If 4 2, 8 4 2 4 2

x y

x y

= + = − + = −

= − = − − = +

Solutions: ( ) ( )4 2, 4 2 and 4 2, 4 2+ − − +

8. 24

2 4

y x

y x

= −= +

(–2, 0) and (–1.2, 1.6) are the intersection points.


2 2

2

4 2 4

4 4 16 16

5 16 12 0

( 2)(5 6) 0

62 or

58

0 or 5

x x

x x x

x x

x x

x x

y y

− = +− = + +

+ + =+ + =

= − = −

= =

Solutions: ( ) 6 82, 0 and ,

5 5− −

9. 2

y x

y x

== −

(1, 1) is the intersection point.

Solve by substitution:

2

2

2

4 4

5 4 0

( 4)( 1) 0

4 or 1

2 or =1

x x

x x x

x x

x x

x x

y y

= −= − +

− + =− − =

= == −

Eliminate (4, –2); we must have 0y ≥ . Solution: (1, 1)


1115


10. 6

y x

y x

== −



2

2

6

36 12

13 36 0

( 4)( 9) 0

4 or 9

2 or 3

x x

x x x

x x

x x

x x

y y

= −= − +

− + =− − =

= == = −

Eliminate (9, –3); we must have 0y ≥ . Solution: (4, 2)

11. 2

2

2

x y

x y y

== −



2

2 2

4 0

( 4) 0

0 or =4

0 or =8

y y y

y y

y y

y y

x x

= −− =− =

==


12. 2

1

6 9

y x

y x x

= −= − +



2

6 9 1

7 10 0

( 2)( 5) 0

2 or =5

1 or =4

x x x

x x

x x

x x

y y

− + = −− + =

− − ===


13. 2 2

2 2

4

2 0

x y

x x y

+ =+ + =

(–2, 0) is the intersection point.

Substitute 4 for 2 2x y+ in the second equation.

2

2 4 0

2 4

2

4 ( 2) 0

x

x

x

y

+ == −= −

= − − =

Solution: (–2, 0)


1116


14. 2 2

2 2

8

4 0

x y

x y y

+ =+ + =

(–2, –2) and (2, –2) are the intersection points.

Substitute 8 for 2 2x y+ in the second equation.

2

8 4 0

4 8

2

8 ( 2) 2

y

y

y

x

+ == −= −

= ± − − = ±

Solution: (–2, –2) and (2, –2)

15. 2 2

3 5

5

y x

x y

= −+ =

(1, –2) and (2, 1) are the intersection points.

Solve by substitution: 2 2

2 2

2

2

(3 5) 5

9 30 25 5

10 30 20 0

3 2 0

( 1)( 2) 0

x x

x x x

x x

x x

x x

+ − =+ − + =

− + =− + =

− − =

1 or 2

2 1

x x

y y

= == − =

Solutions: (1, –2) and (2, 1)

16. 2 2 10

2

x y

y x

+ == +

(1, 3) and (–3, –1) are the intersection points.

Solve by substitution: 2 2

2 2

2

( 2) 10

4 4 10

2 4 6 0

2( 3)( 1) 0

x x

x x x

x x

x x

+ + =+ + + =

+ − =+ − =

3 or 1

1 3

x x

y y

= − == − =

Solutions: (–3, –1) and (1, 3)

17. 2 2

2

4

4

x y

y x

+ =− =

(–1, 1.73), (–1, –1.73), (0, 2), and (0, –2) are the intersection points.

2Substitute 4 for in the first equation:x y+ 2

2

2 2

4 4

0

( 1) 00 or 1

4 3

2 3

x x

x x

x xx x

y y

y y

+ + =+ =+ =

= = −= == ± = ±

Solutions: ( ) ( )(0, – 2), (0, 2), 1, 3 , 1, 3− − −


1117


18. 2 2

2

16

2 8

x y

x y

+ =− =

(–3.46, 2), (0, –4), and (3.46, 2) are the intersection points.

Substitute 2 8y + for 2x in the first equation. 2

2

2 8 16

2 8 0

( 4)( 2) 0

y y

y y

y y

+ + =+ − =

+ − =

2 2

4 or 2

0 or 12

0 2 3

y y

x x

x x

= − == =

= = ±

Solutions: ( ) ( )(0, – 4), 2 3, 2 , 2 3, 2−

19. 2 2

4

8

xy

x y

=+ =

(–2, –2) and (2, 2) are the intersection points.


2

22

4 2

4 2

2 2

2

2

48

168

16 88 16 0

( 4) 0

4 042 or 22 or 2

xx

xx

x xx x

x

xxx xy y

+ =

+ =

+ =− + =

− =− =

== = −= = −

Solutions: (–2, –2) and (2, 2)

20. 2

1

x y

xy

==



3

2

1

1

1

(1) 1

xx

x

x

y

=

=== =

Solution: (1, 1)

21. 2 2

2

4

9

x y

y x

+ == −

No solution; Inconsistent.

Solve by substitution: 2 2 2

2 4 2

4 2

( 9) 4

18 81 4

17 77 0

x x

x x x

x x

+ − =+ − + =

− + =

2 17 289 4(77)

2

17 192

x± −

=

± −=

There are no real solutions to this expression. Inconsistent.


1118


22. 1

2 1

xy

y x

== +

(–1, –1) and (0.5, 2) are the intersection points.


2

(2 1) 1

2 1 0

( 1)(2 1) 0

11 or

21 2

x x

x x

x x

x x

y y

+ =+ − =

+ − =

= − =

= − =

Solutions: (–1, –1) and 1

, 22

23. 2 4

6 13

y x

y x

= −= −



2

2

2

4 6 13

6 9 0

( 3) 0

3 0

3

(3) 4 5

x x

x x

x

x

x

y

− = −− + =

− =− =

== − =

Solution: (3,5)

24. 2 2 10

3

x y

xy

+ ==

(1, 3), (3, 1), (–3, –1), and (–1, –3) are the intersection points.


( )22

2

4 2

4 2

2 2

2

3

9

10

10

9 10

10 9 0

( 9)( 1) 0

( 3)( 3)( 1)( 1) 0

x

x

x

x

x x

x x

x x

x x x x

+ =

+ =

+ =− + =

− − =− + − + =

3 or –3 or 1 or 1

1 1 3 –3

x x x x

y y y y

= = = = −= = − = =

Solutions: (3, 1), (–3, –1), (1, 3), (–1, –3)

25. Solve the second equation for y, substitute into the first equation and solve:

2 22 18

44

x y

xy yx

+ =

= =

( )( )

22

22

4 2

4 2

4 2

2 2

42 18

162 18

2 16 18

2 18 16 0

9 8 0

8 1 0

xx

xx

x x

x x

x x

x x

+ =

+ =

+ =− + =

− + =

− − =

2 28 or 1or 18= 2 2

x x

xx

= == ±= ± ±


1119


4If 2 2 : 2

2 24

If 2 2 : 22 2

4If 1: 4

14

If 1: 41

x y

x y

x y

x y

= = =

= − = = −−

= = =

= − = = −−

Solutions:

( ) ( )2 2, 2 , 2 2, 2 , (1, 4), ( 1, 4)− − − −


2 2 21

7 7

x y

x y y x

− =+ = = −

( )( )

22

2 2

7 21

49 14 21

14 70

5

7 5 2

x x

x x x

x

x

y

− − =

− − + =

=== − =

Solution: (5, 2)

27. Substitute the first equation into the second equation and solve:

2 2

2 1

2 1

y x

x y

= ++ =

( )

( )

22

2 2

2

2 2 1 1

2 4 4 1 1

6 4 0

2 3 2 0

x x

x x x

x x

x x

+ + =

+ + + =+ =+ =

2 0 or 3 2 0

20 or

3

x x

x x

= + =

= = −

If 0 : 2(0) 1 1

2 2 4 1If : 2 1 1

3 3 3 3

x y

x y

= = + =

= − = − + = − + = −

Solutions: 2 1

(0, 1), ,3 3

− −

28. Solve the second equation for x and substitute into the first equation and solve:

2 24 162 2 2 2

x yy x x y

− =− = = −

( )

2 2

2 2

32

(2 2) 4 16

4 8 4 4 168 12

32

2 2 5

y y

y y yy

y

x

− − =− + − =

− =

= −

= − − = −

Solutions: ( )325,− −

29. Solve the first equation for y, substitute into the second equation and solve:

2 2

1 0 1

6 5

x y y x

x y y x

+ + = = − −+ + − = −

2 2

2 2

2

( 1) 6( 1) 5

2 1 6 6 5

2 5 0(2 5) 0

x x x x

x x x x x

x xx x

+ − − + − − − = −+ + + − − − = −

− =− =

520 or x x= =

If 0 : (0) 1 15 5 7

If : 12 2 2

x y

x y

= = − − = −

= = − − = −

Solutions: ( )5 72 2(0, 1), ,− −


2 22 84

4

x xy y

xy yx

− + =

= =

( ) ( )( )( )

22

22

4 2

4 2

2 2

4 42 8

162 4 8

2 16 12

6 8 0

4 2 0

( 2)( 2) 2 2 0

x xx x

xx

x x

x x

x x

x x x x

− + =

− + =

+ =− + =

− − =

− + − + =

2 or 2 or 2 or 2

2 2 2 2 2 2

x x x x

y y y y

= = − = = −= = − = = −

Solutions:

( ) ( )( 2, 2), 2, 2 2 , 2, 2 2 , (2, 2)− − − −


1120



2 24 3 9 15

2 52 3 5

3 3

x xy y

x y y x

− + =

+ = = − +

2

2

2 2 2

2

2 5 2 54 3 9 15

3 3 3 3

4 2 5 4 20 25 15

10 25 10 0

x x x x

x x x x x

x x

− − + + − + =

+ − + − + =− + =

22 5 2 0

(2 1)( 2) 0

x x

x x

− + =− − =

1

or 22

x x= =

1 2 1 5 4If :

2 3 2 3 32 5 1

If 2 : (2)3 3 3

x y

x y

= = − + =

= = − + =

Solutions: 1 4 1

, , 2,2 3 3

32. 22 3 6 2 4 0

2 3 4 0

y xy y x

x y

− + + + =− + =

Solve the second equation for x, substitute into the first equation and solve: 2 3 4 0

2 3 4

3 42

x y

x y

yx

− + == −

−=

2

2 2

2

2

3 4 3 42 3 6 2 4

2 29

2 6 6 3 4 42

515 0

25 30 0

5 ( 6) 0

y yy y y

y y y y y

y y

y y

y y

− −− + + = −

− + + + − = −

− + =

− + =− − =

0 or 6y y= =

( )3 0 4If 0 : 2

2y x

−= = −

( )3 6 4If 6 : 7

2y x

−= = =

Solutions: (–2, 0), (7, 6)

33. Multiply each side of the second equation by 4 and add the equations to eliminate y:

2 2 2 2

42 2 2 2

2

2

4 7 4 7

3 31 12 4 124

13 117

9 3

x y x y

x y x y

x

xx

− = − → − = −+ = → + =

=== ±

If 3x = : 2 2 23(3) 31 4 2y y y+ = = = ±

If 3x = − : 2 2 23( 3) 31 4 2y y y− + = = = ± Solutions: (3, 2), (3, –2), (–3, 2), (–3, –2)

34. 2 2

2 2

3 2 5 0

2 2 0

x y

x y

− + =− + =

2 2

2 2

3 2 5

2 2

x y

x y

− = −− = −

Multiply each side of the second equation by –2 and add the equations to eliminate y:

2 2

2 2

2

2

3 2 5

4 2 4

1

1 1

x y

x y

x

xx

− = −− + =

− = −== ±

2 2 2

2 2 2

If 1:

2(1) 2 4 2

If 1:

2( 1) 2 4 2

x

y y y

x

y y y

=− = − = = ±

= −− − = − = = ±

Solutions: (1, 2), (1, –2), (–1, 2), (–1, –2)

35. 2 2

2 2

7 3 5 0

3 5 12

x y

x y

− + =+ =

2 2

2 2

7 3 5

3 5 12

x y

x y

− = −+ =

Multiply each side of the first equation by 5 and each side of the second equation by 3 and add the equations to eliminate y:

2 2

2 2

2

2

35 15 25

9 15 36

44 11

1

41

2

x y

x y

x

x

x

− = −+ =

=

=

= ±


1121


22 2

22 2

12

12

If :

1 9 33 5 12

2 4 2

If :

1 9 33 5 12

2 4 2

x

y y y

x

y y y

=

+ = = = ±

= −

− + = = = ±

Solutions: 1 3 1 3 1 3 1 3

, , , , , , ,2 2 2 2 2 2 2 2

− − − −

36. 2 2

2 2

3 1 0

2 7 5 0

x y

x y

− + =− + =

2 2

2 2

3 1

2 7 5

x y

x y

− = −− = −

Multiply each side of the first equation by –2 and add the equations to eliminate x:

2 2

2 2

2

2

2 6 2

2 7 5

3

3

3

x y

x y

y

y

y

− + =− = −

− = −=

= ±

( )

( )

22 2

22 2

If 3 :

3 3 1 8 2 2

If 3 :

3 3 1 8 2 2

y

x x x

y

x x x

=

− = − = = ±

= −

− − = − = = ±

Solutions:

( ) ( ) ( )( )2 2, 3 , 2 2, 3 , 2 2, 3 ,

2 2, 3

− −

− −

37. Multiply each side of the second equation by 2 and add the equations to eliminate xy:

2 2

22 2

2

2

2 10 2 10

3 2 6 2 4

7 14

2

2

x xy x xy

x xy x xy

x

x

x

+ = → + =

− = → − =

==

= ±

( )

( ) ( )

2

2

If 2 :

3 2 2 2

42 4 2 2

2

If 2 :

3 2 2 2

42 4 2 2

2

x

y

y y y

x

y

y y y

=

− ⋅ =

− ⋅ = − = =

= −

− − − =

−⋅ = − = = −

Solutions: ( ) ( )2, 2 2 , 2, 2 2− −

38. 2

2

5 13 36 0

7 6

xy y

xy y

+ + =+ =

2

2

5 13 36

7 6

xy y

xy y

+ = −+ =

Multiply each side of the second equation by –5 and add the equations to eliminate xy:

2

2

2

2

5 13 36

5 35 30

22 66

3

3

xy y

xy y

y

y

y

+ = −− − = −

− = −=

= ±

( ) ( )2

If 3 :

3 7 3 6

153 15 5 3

3

y

x

x x x

=

+ =

−⋅ = − = = −

( ) ( )2

If 3 :

3 7 3 6

153 15 5 3

3

y

x

x x x

= −

− + − =

− ⋅ = − = =

Solutions: ( ) ( )5 3, 3 , 5 3, 3− −


1122


39. 2 2

2 2

2 2

2 8 0

x y

x y

+ =− + =

2 2

2 2

2 2

2 8

x y

x y

+ =− = −


2 2

2 2

2

2

4 2 4

2 8

5 44

5

x y

x y

x

x

+ =− = −

= −

= −

No real solution. The system is inconsistent.

40. 2 2

2 2

4 0

2 3 6

y x

x y

− + =+ =

2 2

2 2

4

2 3 6

x y

x y

− =+ =

Multiply each side of the first equation by 2− and add the equations to eliminate x:

2 2

2 2

2

2

2 2 8

2 3 6

5 22

5

x y

x y

y

y

− + = −+ =

= −

= −

No real solution. The system is inconsistent.

41. 2 2

2 2

2 16

4 24

x y

x y

+ =− =

Multiply each side of the second equation by 2 and add the equations to eliminate y:

2 2

2 2

2

2

2 16

8 2 48

9 6464

98

3

x y

x y

x

x

x

+ =− =

=

=

= ±

22 2

2

8If :

38 80

2 16 23 9

40 2 109 3

x

y y

y y

=

+ = =

= = ±

22 2

2

8If :

38 80

2 16 23 9

40 2 109 3

x

y y

y y

= −

− + = =

= = ±

Solutions:

8 2 10 8 2 10 8 2 10, , , , , ,

3 3 3 3 3 3

8 2 10,

3 3

− −

− −

42. 2 2

2 2

4 3 4

2 6 3

x y

x y

+ =− = −


2 2

2 2

2

2

8 6 8

2 6 3

10 5

1

2

2

2

x y

x y

x

x

x

+ =− = −

=

=

= ±

22 2

2

22 2

2

2If :

2

24 3 4 3 2

2

2 63 3

2If :

2

24 3 4 3 2

2

2 63 3

x

y y

y y

x

y y

y y

=

+ = =

= = ±

= −

− + = =

= = ±

Solutions:

2 6 2 6 2 6, , , , , ,

2 3 2 3 2 3

2 6,

2 3

− −

− −


1123


43. 2 2

2 2

5 23 0

3 17

x y

x y

− + =

+ =

2 2

2 2

5 23

3 17

x y

x y

− = −

+ =

Multiply each side of the second equation by 2 and add the equations to eliminate y:

2 2

2 2

2

2

5 23

6 214

11 11

1

1

x y

x y

x

x

x

− = −

+ =

=

== ±

22 2 2

22 2 2

If 1:

3 1 1 17 4

4(1)

12

If 1:

3 1 1 17 4

4( 1)

12

x

yy y

y

x

yy y

y

=

+ = = =

= ±

= −

+ = = =−

= ±

Solutions: 1 1 1 1

1, , 1, , 1, , 1,2 2 2 2

− − − −

44. 2 2

2 2

2 31 0

6 72 0

x y

x y

− + =

− + =

2 2

2 2

2 31

6 72

x y

x y

− = −

− = −


2 2

2 2

2

2

6 93

6 72

2 1

2

2

y y

x y

y

y

y

− + =

− = −

=

=

= ±

( )

( )

2 2 2

2

2 2 2

2

2 3 2 1If 2 : 1

22

4 2

2 3 2 1If 2 : 1

22

4 2

yx x

x x

yx x

x x

= − = − =

= = ±

= − − = − =−

= = ±

Solutions:

( ) ( ) ( ) ( )2, 2 , 2, 2 , 2, 2 , 2, 2− − − −

45. 4 4

4 4

1 66

2 219

x y

x y

+ =

− =


4 4

4 4

4

4

2 1212

2 219

147

2

x y

x y

y

y

− − = −

− =

− =

= −

There are no real solutions. The system is inconsistent.


1124


46. Add the equations to eliminate y:

4 4

4 4

4

4

4

1 11

1 14

2 5

2

52

5

x y

x y

x

x

x

− =

+ =

=

=

= ±

44 4 4

4

4 4

44 4 4

4

4 4

2 1 1 1 3If : 4

5 225

2 23 3

2 1 1 1 3If : 4

5 225

2 23 3

xy y

y y

xy y

y y

= + = =

= = ±

= − + = =

−

= = ±

Solutions:

4 4 4 4 4 4

4 4

2 2 2 2 2 2, , , , , ,

5 3 5 3 5 3

2 2,

5 3

− −

− −

47. 2 2

2

3 2 0

6

x xy y

x xy

− + =+ =

Subtract the second equation from the first to eliminate the 2x term.

2

2

4 2 6

2 3

xy y

xy y

− + = −− =

Since 0y ≠ , we can solve for x in this equation to get

2 32

yx

y+= , 0y ≠

Now substitute for x in the second equation and solve for y.

2

22 2

4 2 2

2

6

3 36

2 2

6 9 36

24

x xy

y yy

y y

y y y

y

+ =

+ ++ =

+ + ++ =

( )( )

4 2 4 2 2

4 2

4 2

2 2

6 9 2 6 24

3 12 9 0

4 3 0

3 1 0

y y y y y

y y

y y

y y

+ + + + =− + =− + =

− − =

Thus, 3y = ± or 1y = ± . If 1: 2 1 2

If 1: 2( 1) 2

y x

y x

= = ⋅ == − = − = −

If 3 : 3

If 3 : 3

y x

y x

= =

= − = −

Solutions: ( ) ( )(2, 1), (–2, 1), 3, 3 , 3, 3− − −

48. 2 22 0

6 0

x xy y

xy x

− − =+ + =

Factor the first equation, solve for x, substitute into the second equation and solve:

2 22 0

( 2 )( ) 0

2 or

x xy y

x y x y

x y x y

− − =− + =

= = −

Substitute 2x y= and solve:

2

2

6 0

(2 ) 2 6

2 2 6 0

2( 3) 0

xy x

y y y

y y

y y

+ + =+ = −

+ + =+ + =

21 1 4(1)(3) (No real solution)

2(1)y

− ± −=

Substitute x y= − and solve:

2

6 0

( ) 6

6 0

( 3)( 2) 0

3 or =2

If 3 : 3

If 2 : 2

xy x

y y y

y y

y y

y y

y x

y x

+ + =− ⋅ + − = −− − + =

− − − == −= − == = −

Solutions: (3, –3), (– 2, 2)


1125


49. 2 2 2 0

21 0

y y x x

xy

y

+ + − − =−+ + =

Multiply each side of the second equation by –y and add the equations to eliminate y:

( )

2 2

2

2

2 0

2 0

2 0

2 0

y y x x

y y x

x x

x x

+ + − − =− − − + =

− =− =

0 or 2x x= =

2 2 2

2 2 2

If 0 :

0 0 2 0 2 0

( 2)( 1) 0 2 or 1

If 2 :

2 2 2 0 0

( 1) 0 0 or 1

Note: 0 because of division by zero.

x

y y y y

y y y y

x

y y y y

y y y y

y

=+ + − − = + − =

+ − = = − =

=+ + − − = + =

+ = = = −≠

Solutions: (0, –2), (0, 1), (2, –1)

50. 3 2 2

2

2

2 3 4 0

2 0

x x y y

y yx

x

− + + − =−− + =

Multiply each side of the second equation by 2x− and add the equations to eliminate x: 3 2 2

3 2 2

2 3 4 0

2 0

4 4 0

4 4

1

x x y y

x x y y

y

y

y

− + + − =− + − + =

− ===

3 2 2 3 2

2

If 1:

2 1 3 1 4 0 2 0

( 2) 0 0 or 2

Note: 0 because of division by zero.

y

x x x x

x x x x

x

=− + + ⋅ − = − =

− = = =≠

Solution: (2, 1)

51. Rewrite each equation in exponential form: 3

5

log 3

log (4 ) 5 4

x

x

y y x

y y x

= → =

= → =

Substitute the first equation into the second and solve:

3 5

5 3

3 2

4

4 0

( 4) 0

x x

x x

x x

=− =

− =

3 20 or 4 0 or 2x x x x= = = = ± The base of a logarithm must be positive, thus

0 and 2x x≠ ≠ − .

3If 2 : 2 8x y= = = Solution: (2, 8)

52. Rewrite each equation in exponential form: 3

2

log (2 ) 3 2

log (4 ) 2 4

x

x

y y x

y y x

= =

= =

Multiply the first equation by 2 then substitute the first equation into the second and solve:

3 2

3 2

2

2

2 0

(2 1) 0

x x

x x

x x

=− =− =

2 1 10 or or 0

2 2x x x x= = = =

The base of a logarithm must be positive, thus 0x ≠ .

21 1 1 1If : 4

2 2 4 16x y y= = = =

Solution: 1 1

,2 16

53. Rewrite each equation in exponential form: 44ln ln 4ln 4ln y yx y x e e y= = = =

23 3 3

3 3

2 2log 2log log2 2 2

log 2 2log

3 3 3 3 3 9y y y

x y

x y+

= +

= = ⋅ = ⋅ =

So we have the system 4

29

x y

x y

==

Therefore we have : 2 4 2 4 2 29 9 0 (9 ) 0y y y y y y= − = − =

2 (3 )(3 ) 0

0 or 3 or 3

y y y

y y y

+ − == = − =

Since ln y is undefined when 0y ≤ , the only


1126


solution is 3y = . 4 4If 3 : 3 81y x y x= = = =

Solution: ( )81, 3

54. Rewrite each equation in exponential form: 55ln ln 5ln 5ln y yx y x e e y= = = =

22 2 2

2 2

3 2log 2log log3 3 2

log 3 2log

2 2 2 2 2 8y y y

x y

x y+

= +

= = ⋅ = ⋅ =

So we have the system 5

28

x y

x y

==

Therefore we have

( )

2 5

2 5

2 3

8

8 0

8 0

y y

y y

y y

=− =

− =

30 or 8 0 2y y y= − = =

Since ln y is undefined when 0y ≤ , the only solution is 2y = .

5 5If 2 : 2 32y x y x= = = =

Solution: ( )32, 2

55.

( ) ( )( ) ( )

2 2

2

22 31 12 2 2

2 21 1 12 2 2

3 2 0

1 0

x x y y

y yx

x

x y

x y

+ + − + =−+ + =

+ + − =

+ + − =

56.

( ) ( )( )

2 2

2 2 51 12 2 2

2 912 4

2 0

21 0

y y x x

xy

y

x y

x y

+ + − − =−+ + =

− + + =

= − + +

57. Graph: 1 2(2 / 3); ( )y x y e x= ∧ = ∧ − Use INTERSECT to solve:

4.7–4.7

3.1

-3.1 Solution: 0.48, 0.62x y= = or (0.48, 0.62)

58. Graph: 1 2(3 / 2); ( )y x y e x= ∧ = ∧ − Use INTERSECT to solve:

4.7–4.7

3.1

–3.1 Solution: 0.65, 0.52x y= = or (0.65, 0.52)

59. Graph: 3 2 31 22 ; 4 /y x y x= − =

Use INTERSECT to solve:


1127


4.7–4.7

3.1

–3.1 Solution: 1.65, 0.89x y= − = − or (–1.65, –0.89)

60. Graph: 3 31 2

23

2 ; 2 ;

4 /

y x y x

y x

= − = − −

=


4.7–4.7

3.1

–3.1 Solution: 1.37, 2.14x y= − = or (–1.37, 2.14)

61. Graph: 4 44 41 2

3 4

12 ; 12 ;

2 / ; 2 /

y x y x

y x y x

= − = − −

= = −


Solutions: 0.58, 1.86;x y= = 1.81, 1.05;x y= =

1.81, 1.05;x y= = − ; 0.58, 1.86x y= = − or (0.58, 1.86), (1.81, 1.05), (1.81, –1.05), (0.58, –1.86)

62. Graph: 4 44 41 2

3

6 ; 6 ;

1/

y x y x

y x

= − = − −=


Solutions: 0.64, 1.55;x y= = 1.55, 0.64;x y= =

0.64, 1.55;x y= − = − ; 1.55, 0.64x y= − = − or (0.64, 1.55), (1.55, 0.64), (–0.64, –1.55), (–1.55, –0.64)

63. Graph: 1 22 / ; lny x y x= = Use INTERSECT to solve:

4.7–4.7

3.1

–3.1 Solution: 2.35, 0.85x y= = or (2.35, 0.85)


1128


64. Graph: 2 21 2

3

4 ; 4 ;

ln

y x y x

y x

= − = − −=


Solution: 1.90, 0.64;x y= = 0.14, 2.00x y= = − or (1.90, 0.64), (0.14, –2.00)

65. Solve the first equation for x, substitute into the second equation and solve:

2 2

2 0 2

( 1) ( 1) 5

x y x y

x y

+ = = −− + − =

2 2

2 2 2

( 2 1) ( 1) 5

4 4 1 2 1 5 5 2 3 0

(5 3)( 1) 0

3=0.6 or 1

56

= 1.2 or 25

y y

y y y y y y

y y

y y

x x

− − + − =+ + + − + = + − =

− + =

= = −

= − − =

The points of intersection are 6 3

, , (2, –1)5 5

− .


2 2

2 6 0 2 6

( 1) ( 1) 5

x y x y

x y

+ + = = − −+ + + =

2 2

2 2

2

( 2 6 1) ( 1) 5

4 20 25 2 1 5

5 22 21 0

(5 7)( 3) 0

y y

y y y y

y y

y y

− − + + + =+ + + + + =

+ + =+ + =

7

or 3516

or 05

y y

x x

= − = −

= − =

The points of intersection are 16 7

, , (0, 3)5 5

− − − .

67. Complete the square on the second equation.

2

2

4 4 1 4

( 2) 3

y y x

y x

+ + = − ++ = +

Substitute this result into the first equation. 2

2

2

( 1) 3 4

2 1 3 4

0

( 1) 0

0 or 1

x x

x x x

x x

x x

x x

− + + =− + + + =

− =− =

= =2

2

If 0 : ( 2) 0 3

2 3 2 3

If 1: ( 2) 1 3

2 2 2 2

x y

y y

x y

y y

= + = +

+ = ± = − ±= + = +

+ = ± = − ±

The points of intersection are:

( ) ( ) ( ) ( )0, 2 3 , 0, 2 3 , 1, – 4 , 1, 0− − − + .


1129


68. Complete the square on the second equation, substitute into the first equation and solve:

2 2

2

( 2) ( 1) 4

2 5 0

x y

y y x

+ + − =− − − =

2

2

2 1 5 1

( 1) 6

y y x

y x

− + = + +− = +

2

2

2

( 2) 6 4

4 4 6 4

5 6 0

( 2)( 3) 0

2 or 3

x x

x x x

x x

x x

x x

+ + + =+ + + + =

+ + =+ + =

= − = −

2

2

If 2 : ( 1) 2 6 1 2

1 or 3

If 3 : ( 1) 3 6 1 3

1 3

x y y

y y

x y y

y

= − − = − + − = ±= − =

= − − = − + − = ±

= ±

The points of intersection are:

( ) ( )3, 1 3 , 3,1 3 , ( 2, 1), ( 2, 3)− − − + − − − .


2 2

43

6 1 0

yx

x x y

=−

− + + =

43

43

43

yx

xy

xy

=−

− =

= +

22

22

22

4 43 6 3 1 0

16 24 249 18 1 0

168 0

yy y

yy yy

yy

+ − + + + =

+ + − − + + =

+ − =

4 2

4 2

2 2

2

2

16 8 0

8 16 0

( 4) 0

4 0

4

2

y y

y y

y

y

y

y

+ − =− + =

− =− =

== ±

4If 2 : 3 5

24

If 2 : 3 12

y x

y x

= = + =

= − = + =−

The points of intersection are: (1, –2), (5, 2).


1130


70. Substitute the first equation into the second equation and solve:

2 2

42

4 4 0

yx

x x y

=+

+ + − =

( )( )( )

( )

22

22

2 2

2 2

4 3 2

4 3 2

2 2

2 2

44 4 0

2

44 4

2

( 2) 4 4 16

4 4 4 4 16

8 16 16 16

8 16 0

8 16 0

( 4) 0

0 or 4

2 2

x xx

x xx

x x x

x x x x

x x x

x x x

x x x

x x

x x

y y

+ + − =+

+ − = −+

+ + − = −

+ + + − = −

+ + − = −+ + =

+ + =

+ == = −= = −

The points of intersection are: (0, 2), (–4, –2).

71. Let and x y be the two numbers. The system of equations is:

2 2

2 2

10

x y x y

x y

− = = ++ =


( )

( )( )

2 2

2 2

2

2 10

4 4 10

2 3 0

3 1 0 3 or 1

y y

y y y

y y

y y y y

+ + =

+ + + =+ − =

+ − = = − =

If 3 : 3 2 1

If 1: 1 2 3

y x

y x

= − = − + = −= = + =

The two numbers are 1 and 3 or –1 and –3.


2 2

7 7

21

x y x y

x y

+ = = −− =


( )2 2

2 2

7 21

49 14 21

14 28

2 7 2 5

y y

y y y

y

y x

− − =

− + − =− = −

= = − =

The two numbers are 2 and 5.


2 2

44

8

xy xy

x y

= =

+ =


( )

22

22

4 2

4 2

22

2

48

168

16 8

8 16 0

4 0

4

2

yy

yy

y y

y y

y

y

y

+ =

+ =

+ =− + =

− =

== ±

4 4If 2 : 2; If 2 : 2

2 2y x y x= = = = − = = −

−



2 2

1010

21

xy xy

x y

= =

− =



1131


( )( )

22

22

4 2

4 2

2 2

1021

10021

100 21

21 100 0

4 25 0

yy

yy

y y

y y

y y

− =

− =

− =+ − =

− + =

2 4

2

y

y

== ±

or 2 25 (no real solution)y = −

10210

2

If 2 : 5

If 2 : 5

y x

y x −

= = =

= − = = −



1 15

x y xy

x y

− =

+ =


( )1 1

x xy y

yx y y x

y

− =

− = =−

1

1 13 3

1 23 3

1 15

1 15

25

2 5

6 2

1 1

3 21

yy

y

yy y

yy

y y

y

y x

−+ =

− + =

− =

− ==

= = = =−

The two numbers are 12 and 1

3 .


1 13

x y xy

x y

+ =

− =

Solve the first equation for x, substitute into the

second equation and solve:

( )1 1

xy x y

yx y y x

y

− =

− = =−

1

1 13

1 13

23

2 3

2 2

1 11

1 1 2

yy

y

yy y

yy

y y

y

y x

−− =

− − =

− =

− == −

−= − = =− −

The two numbers are 1− and 12 .

77. 2310 10

ab

a b a b

=

+ = = −

Solve the second equation for a , substitute into the first equation and solve:

10 23

3(10 ) 2

30 3 2

30 5

6 4

bb

b b

b b

b

b a

− =

− =− =

== =

10; 2a b b a+ = − =

The ratio of a b+ to b a− is 102 5= .

78. 4314 14

ab

a b a b

=

+ = = −

Solve the second equation for a , substitute into the first equation and solve:

( )

14 43

3 14 4

42 3 4

42 7

6 8

bb

b b

b b

b

b a

− =

− =− =

== =

2; 14a b a b− = + =

The ratio of a b− to a b+ is 2 1714 = .


1132


79. Let x = the width of the rectangle. Let y = the length of the rectangle.

2 2 16

15

x y

xy

+ ==

Solve the first equation for y, substitute into the second equation and solve. 2 2 16

2 16 2

8

x y

y x

y x

+ == −= −

( )

( )( )

2

2

8 15

8 15

8 15 0

5 3 0

x x

x x

x x

x x

− =

− =− + =

− − =

5x = or 3x = The dimensions of the rectangle are 3 inches by 5 inches.

80. Let 2x = the side of the first square. Let 3x = the side of the second square.

( ) ( )2 2

2 2

2

2

2 3 52

4 9 52

13 52

4

2

x x

x x

x

x

x

+ =

+ ==== ±

Note that we must have 0x > . The sides of the first square are (2)(2) = 4 feet and the sides of the second square are (3)(2) = 6 feet.

81. Let x = the radius of the first circle. Let y = the radius of the second circle.

2 2

2 2 12

20

x y

x y

π + π = ππ + π = π

Solve the first equation for y, substitute into the second equation and solve: 2 2 12

6

6

x y

x y

y x

π + π = π+ =

= −

2 2

2 2

2 2

2 2 2

2

20

20

(6 ) 20

36 12 20 2 12 16 0

6 8 0 ( 4)( 2) 0

4 or 2

2 4

x y

x y

x x

x x x x x

x x x x

x x

y y

π + π = π+ =

+ − =+ − + = − + =

− + = − − == == =

The radii of the circles are 2 centimeters and 4 centimeters.

82. Let x = the length of each of the two equal sides in the isosceles triangle. Let y = the length of the base. The perimeter of the triangle: 18x x y+ + = Since the altitude to the base y is 3, the Pythagorean theorem produces another equation.

2 22 2 23 9

2 4y y

x x+ = + =

Solve the system of equations:

22

2 18 18 2

94

x y y x

yx

+ = = −

+ =

Solve the first equation for y, substitute into the second equation and solve.

( )

( )

22

22

2 2

18 29

4324 72 4

94

81 18 9

18 90

5 18 2 5 8

xx

x xx

x x x

x

x y

−+ =

− + + =

− + + =− = −

= = − =

The base of the triangle is 8 centimeters.

83. The tortoise takes 9 + 3 = 12 minutes or 0.2 hour longer to complete the race than the hare. Let r = the rate of the hare. Let t = the time for the hare to complete the race. Then t + 0.2 = the time for the tortoise and

0.5r − = the rate for the tortoise. Since the length of the race is 21 meters, the distance equations are:

( ) ( )

2121

0.5 0.2 21

r t rt

r t

= =

− + =

Solve the first equation for r, substitute into the


1133


second equation and solve:

( )

( )

( )( )

2

2

210.5 0.2 21

4.221 0.5 0.1 21

4.210 21 0.5 0.1 10 21

210 42 5 210

5 42 0

5 14 3 0

tt

tt

t t tt

t t t t

t t

t t

− + =

+ − − =

+ − − = ⋅

+ − − =+ − =

− + =

5 14 0

5 14

142.8

5

t

t

t

− ==

= =

or 3 0

3

t

t

+ == −

3t = − makes no sense, since time cannot be negative. Solve for r:

217.5

2.8r = =

The average speed of the hare is 7.5 meters per hour, and the average speed for the tortoise is 7 meters per hour.

84. Let 1 2 3, ,v v v = the speeds of runners 1, 2, 3.

Let 1 2 3, ,t t t = the times of runners 1, 2, 3. Then by the conditions of the problem, we have the following system:

1 1

2 1

3 1

2 2

5280

5270

5260

5280

v t

v t

v t

v t

====

Distance between the second runner and the third runner after 2t seconds is:

2 23 2 3 1

2 1

5280 5280

52805280 5260

527010.02

v tv t v t

v t− = −

= −

≈

The second place runner beats the third place runner by about 10.02 feet.

85. Let x = the width of the cardboard. Let y = the length of the cardboard. The width of the box will be 4x − , the length of the box will be

4y − , and the height is 2. The volume is

( 4)( 4)(2)V x y= − − . Solve the system of equations:

216216

2( 4)( 4) 224

xy yx

x y

= =

− − =


( )

( )( )

2

2

2

2162 8 4 224

1728432 8 32 224

432 8 1728 32 224

8 240 1728 0

30 216 0

12 18 0

xx

xx

x x x x

x x

x x

x x

− − =

− − + =

− − + =− + =

− + =− − =

12 0

12

x

x

− ==

or 18 0

18

x

x

− ==

The cardboard should be 12 centimeters by 18 centimeters.

86. Let x = the width of the cardboard. Let y = the length of the cardboard. The area of the cardboard is: 216xy =

The volume of the tube is: 2 224V r h= π =

where and 2 or 2x

h y r x r= π = =π

.

Solve the system of equations:

2 2

216216

224 2242 4

xy yx

x x yy

= =

π = =π π


( )2 216

2244216 896

89613.03

216

xx

x

x

πππ

=

=

= ≈

( )2

896216

216216 21616.57

896y

x π π= = = ≈

The cardboard should be about 13.03 centimeters by 16.57 centimeters.


1134


87. Find equations relating area and perimeter: 2 2 4500

3 3 ( ) 300

x y

x y x y

+ =+ + − =

Solve the second equation for y, substitute into the first equation and solve: 4 2 300

2 300 4

150 2

x y

y x

y x

+ == −= −

2 2

2 2

2

2

2

(150 2 ) 4500

22,500 600 4 4500

5 600 18,000 0

120 3600 0

( 60) 0

60 0

60

150 2(60) 30

x x

x x x

x x

x x

x

x

x

y

+ − =+ − + =

− + =− + =

− =− =

== − =

The sides of the squares are 30 feet and 60 feet.

88. Let x = the length of a side of the square. Let r = the radius of the circle. The area of the square is 2x and the area of the

circle is 2rπ . The perimeter of the square is 4x and the circumference of the circle is 2 rπ . Find equations relating area and perimeter:

2 2 100

4 2 60

x r

x r

+ π =+ π =

Solve the second equation for x, substitute into the first equation and solve: 4 2 60

4 60 2

115

2

x r

x r

x r

+ π == − π

= − π

22

2 2 2

2 2

115 100

21

225 15 1004

115 125 0

4

r r

r r r

r r

− π + π =

− π + π + π =

π + π − π + =

2 2 2

2 2

2

14 ( 15 ) 4 (125)

4

1225 500

4

100 500 0

b ac− = − π − π + π

= π − π + π

= π − π <

Since the discriminant is less than zero, it is impossible to cut the wire into two pieces whose total area equals 100 square feet.

89. Solve the system for and l w : 2 2l w P

l w A

+ ==

Solve the first equation for l , substitute into the second equation and solve.

2

2

2 2

2

2

2

02

l P w

Pl w

Pw w A

Pw w A

Pw w A

= −

= −

− =

− =

− + =

22

2

2

2 42 4

2

164 4

2 2

16162

2 4

P PP P

P P

AA

w

AP P A

± −± −= =

−± ± −= =

2

2 2

2

2 2

16If then

4

16 162 4 4

16If then

4

16 162 4 4

P P Aw

P P P A P P Al

P P Aw

P P P A P P Al

+ −=

+ − − −= − =

− −=

− − + −= − =

If it is required that length be greater than width, then the solution is:

2 216 16and

4 4P P A P P A

w l− − + −= =

90. Solve the system for and l b :

22 2

2 2

4

P b l b P l

bh l

= + = −

+ =

Solve the first equation for b , substitute into the second equation and solve.


1135


( )

2 2 2

22 2

2 2 2 2

2 2

2 2

4 4

4 2 4

4 4 4 4

4 4

44

h b l

h P l l

h P Pl l l

h P Pl

h Pl

P

+ =

+ − =

+ − + =+ =

+=

2 2 2 24 42 2

h P P hb P

P P+ −= − =

91. Solve the equation: 2 4(2 4) 0m m− − =

( )

2

2

8 16 0

4 0

4

m m

m

m

− + =

− ==

Use the point-slope equation with slope 4 and the point (2, 4) to obtain the equation of the tangent line: 4 4( 2) 4 4 8 4 4y x y x y x− = − − = − = −

92. Solve the system: 2 2 10x y

y mx b

+ == +

Solve the system by substitution:

( )

( )

22

2 2 2 2

2 2 2

10

2 10 0

1 2 10 0

x mx b

x m x bmx b

m x bmx b

+ + =

+ + + − =

+ + + − =

Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 (1) 3m b b m= + = − There is one solution to the quadratic if the discriminant is zero.

( ) ( ) ( )2 2 2

2 2 2 2 2 2

2 2

2 4 1 10 0

4 4 40 4 40 0

40 4 40 0

bm m b

b m b m m b

m b

− + − =

− + − + =− + =

Substitute for b and solve:

( )

( )

22

2 2

2

2

2

40 4 3 40 0

40 4 24 36 40 0

36 24 4 0

9 6 1 0

3 1 0

3 1

13

m m

m m m

m m

m m

m

m

m

− − + =

− + − + =+ + =

+ + =

+ == −

= −

1 103 3

3 3b m= − = − − =

The equation of the tangent line is 1 103 3

y x= − + .

93. Solve the system: 2 2y x

y mx b

= += +

Solve the system by substitution: 2 22 2 0x mx b x mx b+ = + − + − =

Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 (1) 3m b b m= + = − Substitute into the quadratic to eliminate b:

2 22 (3 ) 0 ( 1) 0x mx m x mx m− + − − = − + − = Find when the discriminant equals 0:

( ) ( )( )

( )

2

2

2

4 1 1 0

4 4 0

2 0

2 0

2

m m

m m

m

m

m

− − − =

− + =

− =− =

=

3 3 2 1b m= − = − = The equation of the tangent line is 2 1y x= + .

94. Solve the system: 2 5x y

y mx b

+ == +

Solve the system by substitution: 2 25 5 0x mx b x mx b+ + = + + − =

Note that the tangent line passes through (–2, 1). Find the relation between m and b:

( )1 2 2 1m b b m= − + = +

Substitute into the quadratic to eliminate b:


1136


( )2

2

2 1 5 0

2 4 0

x mx m

x mx m

+ + + − =+ + − =

Find when the discriminant equals 0:

( ) ( )( )

( )

2

2

2

4 1 2 4 0

8 16 0

4 0

4 0

4

m m

m m

m

m

m

− − =

− + =

− =− =

=

( )2 1 2 4 1 9b m= + = + =

The equation of the tangent line is 4 9y x= + .

95. Solve the system: 2 22 3 14x y

y mx b

+ == +


( )

( )

22

2 2 2 2

2 2 2

2 3 14

2 3 6 3 14

3 2 6 3 14 0

x mx b

x m x mbx b

m x mbx b

+ + =

+ + + =

+ + + − =

Note that the tangent line passes through (1, 2). Find the relation between m and b: 2 (1) 2m b b m= + = − Substitute into the quadratic to eliminate b:

2 2 2

2 2 2 2

(3 2) 6 (2 ) 3(2 ) 14 0

(3 2) (12 6 ) (3 12 2) 0

m x m m x m

m x m m x m m

+ + − + − − =+ + − + − − =

Find when the discriminant equals 0: 2 2 2 2

2

2

2

(12 6 ) 4(3 2)(3 12 2) 0

144 96 16 0

9 6 1 0

(3 1) 0

3 1 0

13

m m m m m

m m

m m

m

m

m

− − + − − =+ + =

+ + =+ =

+ =

= −

1 72 2

3 3b m= − = − − =


y x= − + .


y mx b

+ == +


( )

( )

22

2 2 2 2

2 2 2

3 7

3 2 7

3 2 7 0

x mx b

x m x mbx b

m x mbx b

+ + =

+ + + =

+ + + − =

Note that the tangent line passes through (–1, 2). Find the relation between m and b: 2 ( 1) 2m b b m= − + = + There is one solution to the quadratic if the discriminant equals 0.

( ) ( )( )2 2 2

2 2 2 2 2 2

2 2

2 2

2 4 3 7 0

4 4 28 12 84 0

28 12 84 0

7 3 21 0

bm m b

b m b m m b

m b

m b

− + − =

− + − + =− + =− + =


( )

( )

22

2 2

2

2

7 3 2 21 0

7 3 12 12 21 0

4 12 9 0

2 3 0

2 3

32

m m

m m m

m m

m

m

m

− + + =

− − − + =− + =

− ==

=

3 72 2

2 2b m= + = + =


y x= + .


y mx b

− == +


( )

( )

22

2 2 2 2

2 2 2

3

2 3

1 2 3 0

x mx b

x m x mbx b

m x mbx b

− + =

− − − =

− − − − =

Note that the tangent line passes through (2, 1). Find the relation between m and b: 1 (2) 1 2m b b m= + = − Substitute into the quadratic to eliminate b:

2 2 2

2 2 2 2

2 2 2 2

(1 ) 2 (1 2 ) (1 2 ) 3 0

(1 ) ( 2 4 ) 1 4 4 3 0

(1 ) ( 2 4 ) ( 4 4 4) 0

m x m m x m

m x m m x m m

m x m m x m m

− − − − − − =− + − + − + − − =

− + − + + − + − =Find when the discriminant equals 0:


1137


( ) ( )( )

( )

22 2 2

2 3 4 4 3

2

2

2

2 4 4 1 4 4 4 0

4 16 16 16 16 16 16 0

4 16 16 0

4 4 0

2 0

2

m m m m m

m m m m m m

m m

m m

m

m

− + − − − + − =

− + − + − + =− + =

− + =

− ==

The equation of the tangent line is 2 3y x= − .

98. Solve the system: 2 22 14y x

y mx b

− == +


( )

( )

2 2

2 2 2 2

2 2 2

2 14

2 4 2 14

2 1 4 2 14 0

mx b x

m x mbx b x

m x mbx b

+ − =

+ + − =

− + + − =

Note that the tangent line passes through (2, 3). Find the relation between m and b: 3 (2) 3 2m b b m= + = − There is one solution to the quadratic if the discriminant equals 0.

( ) ( )( )2 2 2

2 2 2 2 2 2

2 2

2 2

4 4 2 1 2 14 0

16 16 112 8 56 0

112 8 56 0

14 7 0

bm m b

b m b m m b

m b

m b

− − − =

− + + − =+ − =

+ − =


( )

( )

22

2 2

2

2

14 3 2 7 0

14 4 12 9 7 0

18 12 2 0

2 3 1 0

3 1

13

m m

m m m

m m

m

m

m

+ − − =

+ − + − =− + =

− ==

=

1 73 2 3 2

3 3b m= − = − =


y x= + .

99. Solve for 1 2and r r :

1 2

1 2

br r

ac

r ra

+ = −

=

Substitute and solve:

1 2

2 2

22 2 0

br r

ab c

r ra a

b cr r

a a

= − −

− − =

− − − =

22 2 0ar br c+ + =

2

2

1 2

2

2

2

42

42

4 22 2

42

b b acr

ab

r ra

b b ac ba a

b b ac ba a

b b aca

− ± −=

= − − =

− ± −= − −

−= −

− −=

The solutions are: 2 2

1 24 4

and 2 2

b b ac b b acr r

a a− + − − − −= = .


1138


100. Consider the circle with equation

( ) ( )2 2 2x h y k r− + − = and the third degree

polynomial with equation 3 2y ax bx cx d= + + + . Substituting the second equation into the first equation yields

( ) ( )22 3 2 2x h ax bx cx d k r− + + + + − = .

In order to find the roots for this equation we can expand the terms on the left hand side of the

equation. Notice that ( )2x h− yields a 2nd degree

polynomial, and ( )23 2ax bx cx d k+ + + − yields a

6th degree polynomial. Therefore, we need to find the roots of a 6th degree equation, and the Fundamental Theorem of Algebra states that there will be at most six real roots. Thus, the circle and the 3rd degree polynomial will intersect at most six times. Now consider the circle with equation

( ) ( )2 2 2x h y k r− + − = and the polynomial of

degree n with equation 2 3

0 1 2 3 ... nny a a x a x a x a x= + + + + + .

Substituting the first equation into the first equation yields

( ) ( )22 2 3 20 1 2 3 ... n

nx h a a x a x a x a x k r− + + + + + + − =In order to find the roots for this equation we can expand the terms on the left hand side of the equation.

Notice that ( )2x h− yields a 2nd degree polynomial,

and ( )22 30 1 2 3 ... n

na a x a x a x a x k+ + + + + −

yields a polynomial of degree 2n. Therefore, we need to find the roots of an equation of degree 2n, and the Fundamental Theorem of Algebra states that there will be at most 2n real roots. Thus, the circle and the nth degree polynomial will intersect at most 2n times.

101. Since the area of the square piece of sheet metal is 100 square feet, the sheet’s dimensions are 10 feet by 10 feet. Let x = the length of the cut.

10 – 2x 10 – 2x

x

10

10

xx

The dimensions of the box are: length 10 2 ;x= −

width 10 2 ; heightx x= − = . Note that each of these expressions must be positive. So we must have 0 and 10 2 0 5,x x x> − > < that is, 0 5x< < . So the volume of the box is given by

( ) ( ) ( )( )( ) ( )( ) ( )2

length width height

10 2 10 2

10 2

V

x x x

x x

= ⋅ ⋅= − −

= −

a. In order to get a volume equal to 9 cubic feet,

we solve ( ) ( )210 2 9.x x− =

( ) ( )( )

2

2

2 3

10 2 9

100 40 4 9

100 40 4 9

x x

x x x

x x x

− =

− + =

− + =

So we need to solve the equation 3 24 40 100 9 0x x x− + − = .

Graphing 3 21 4 40 100 9y x x x= − + − on a

calculator yields the graph

10−2

−40

80

10−2

−40

80

10−2

−40

80

The graph indicates that there are three real zeros on the interval [0, 6]. Using the ZERO feature of a graphing calculator, we find that the three roots shown occur at 0.093x ≈ ,

4.274x ≈ and 5.632x ≈ . But we’ve already noted that we must have 0< <5x , so the only practical values for the sides of the square base are 0.093x ≈ feet and 4.274x ≈ feet.

b. Answers will vary.

Section 11.7: Systems of Inequalities

1139


Section 11.7

1. 3 4 8

4 4

1

x x

x

x

+ < −<<

{ } ( )1 or ,1x x < −∞

2. 3 2 6x y− = The graph is a line. x-intercept: ( )3 2 0 6

3 6

2

x

x

x

− ===

y-intercept: ( )3 0 2 6

2 6

3

y

y

y

− =− =

= −

3. 2 2 9x y+ = The graph is a circle. Center: (0, 0) ; Radius: 3

4. 2 4y x= + The graph is a parabola. x-intercepts: 2

2

0 4

4, no intercepts

x

x x

= +

= − −

y-intercept: 20 4 4y = + = The vertex has x-coordinate:

( )0

02 2 1b

xa

= − = − = .

The y-coordinate of the vertex is 20 4 4y = + = .

5. True

6. 2y x= ; right; 2

7. dashes; solid

8. half-planes

9. False, see example 9b.

10. unbounded

11. 0x ≥ Graph the line 0x = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (2, 0). Since 2 0 is true, shade the side of the line containing (2, 0).

12. 0y ≥ Graph the line 0y = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 2). Since 2 0 is true, shade the side of the line containing (0, 2).



13. 4x ≥ Graph the line 4x = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (5, 0). Since 5 0 is true, shade the side of the line containing (5, 0).

14. 2y ≤ Graph the line 2y = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (5, 0). Since 0 2 is true, shade the side of the line containing (5, 0).

15. 2 6x y+ ≥ Graph the line 2 6x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6+ ≥ is false, shade the opposite side of the line from (0, 0).

16. 3 2 6x y+ ≤ Graph the line 3 2 6x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≤ is true, shade the side of the line containing (0, 0).

17. 2 2 1x y+ >

Graph the circle 2 2 1x y+ = . Use a dashed line since the inequality uses >. Choose a test point not on the circle, such as (0, 0). Since 2 20 0 1+ > is false, shade the opposite side of the circle from (0, 0).

18. 2 2 9x y+ ≤

Graph the circle 2 2 9x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since

2 20 0 9+ ≤ is true, shade the same side of the circle as (0, 0).


1141


19. 2 1y x≤ −

Graph the parabola 2 1y x= − . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since

20 0 1≤ − is false, shade the opposite side of the parabola from (0, 0).

20. 2 2y x> +

Graph the parabola 2 2y x= + . Use a dashed line since the inequality uses >. Choose a test point not on the parabola, such as (0, 0). Since

20 0 2> + is false, shade the opposite side of the parabola from (0, 0).

21. 4xy ≥ Graph the hyperbola 4xy = . Use a solid line since the inequality uses ≥ . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 4⋅ ≥ is false, shade the opposite side of the hyperbola from (0, 0).

22. 1xy ≤ Graph the hyperbola 1xy = . Use a solid line since the inequality uses ≤ . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 1⋅ ≤ is true, shade the same side of the hyperbola as (0, 0).

23. 2

2 4

x y

x y

+ ≤+ ≥

Graph the line 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 4x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

24. 3 6

2 2

x y

x y

− ≥+ ≤

Graph the line 3 6x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6− ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution.



25. 2 4

3 2 6

x y

x y

− ≤+ ≥ −

Graph the line 2 4x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4− ≤ is true, shade the side of the line containing (0, 0). Graph the line 3 2 6x y+ = − . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≥ − is true, shade the side of the line containing (0, 0). The overlapping region is the solution.

26. 4 5 0

2 2

x y

x y

− ≤− ≥

Graph the line 4 5 0x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (2, 0). Since 4(2) 5(0) 0− ≤ is false, shade the opposite side of the line from (2, 0). Graph the line 2 2x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2− ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

27. 2 3 0

3 2 6

x y

x y

− ≤+ ≤

Graph the line 2 3 0x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 3). Since 2(0) 3(3) 0− ≤ is true, shade the side of the line containing (0, 3). Graph the line 3 2 6x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution.

28. 4 2

2 2

x y

x y

− ≥+ ≥

Graph the line 4 2x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 4(0) 0 2− ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.


1143


29. 2 6

2 4 0

x y

x y

− ≤− ≥

Graph the line 2 6x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6− ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 4 0x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 2). Since 2(0) 4(2) 0− ≥ is false, shade the opposite side of the line from (0, 2). The overlapping region is the solution.

30. 4 8

4 4

x y

x y

+ ≤+ ≥

Graph the line 4 8x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 4(0) 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 4 4x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

31. 2 2

2 2

x y

x y

+ ≥ −+ ≥

Graph the line 2 2x y+ = − . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ − is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

32. 4 4

4 0

x y

x y

− ≤− ≥

Graph the line 4 4x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4− ≤ is true, shade the side of the line containing (0, 0). Graph the line 4 0x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (1, 0). Since 1 4(0) 0− ≥ is true, shade the side of the line containing (1, 0). The overlapping region is the solution.



33. 2 3 6

2 3 0

x y

x y

+ ≥+ ≤

Graph the line 2 3 6x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 0x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 2). Since 2(0) 3(2) 0+ ≤ is false, shade the opposite side of the line from (0, 2). Since the regions do not overlap, the solution is an empty set.

34. 2 0

2 2

x y

x y

+ ≥+ ≥

Graph the line 2 0x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (1, 0). Since 2(1) 0 0+ ≥ is true, shade the side of the line containing (1, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

35. 2 2 9

3

x y

x y

+ ≤+ ≥

Graph the circle 2 2 9x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since

2 20 0 9+ ≤ is true, shade the same side of the circle as (0, 0). Graph the line 3x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 3+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

36. 2 2 9

3

x y

x y

+ ≥+ ≤

Graph the circle 2 2 9x y+ = . Use a solid line since the inequality uses . Choose a test point not on the circle, such as (0, 0). Since

2 20 0 9+ ≥ is false, shade the opposite side of the circle as (0, 0). Graph the line 3x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 3+ ≤ is true, shade the same side of the line as (0, 0). The overlapping region is the solution.


1145


37. 2 4

2

y x

y x

≥ −≤ −

Graph the parabola 2 4y x= − . Use a solid line since the inequality uses . Choose a test point not on the parabola, such as (0, 0). Since

20 0 4≥ − is true, shade the same side of the parabola as (0, 0). Graph the line 2y x= − . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 2≤ − is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

38. 2y x

y x

≤≥

Graph the parabola 2y x= . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (1, 2). Since 22 1≤ is false, shade the opposite side of the parabola from (1, 2). Graph the line y x= . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (1, 2). Since 2 1≥ is true, shade the same side of the line as (1, 2). The overlapping region is the solution.

39. 2 2

2

16

4

x y

y x

+ ≤≥ −

Graph the circle 2 2 16x y+ = . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0,0) . Since

2 20 0 16+ ≤ is true, shade the side of the circle containing (0,0) . Graph the parabola

2 4y x= − . Use a solid line since the inequality is not strict. Choose a test point not on the parabola, such as (0,0) . Since 20 0 4≥ − is true, shade the side of the parabola that contains (0,0) . The overlapping region is the solution.

40. 2 2

2

25

5

x y

y x

+ ≤≤ −

Graph the circle 2 2 25x y+ = . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0,0) . Since

2 20 0 25+ ≤ is true, shade the side of the circle containing (0,0) . Graph the parabola

2 5y x= − . Use a solid line since the inequality is not strict. Choose a test point not on the parabola, such as (0,0) . Since 20 0 5≤ − is false, shade the side of the parabola opposite that which contains the point (0,0) . The overlapping region is the solution.



41. 2

4

1

xy

y x

≥≥ +

Graph the hyperbola 4xy = . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 0 0 4⋅ ≥ is false, shade the opposite side of the hyperbola from (0, 0). Graph the parabola

2 1y x= + . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 20 0 1≥ + is false, shade the opposite side of the parabola from (0, 0).The overlapping region is the solution.

42. 2

2

1

1

y x

y x

+ ≤≥ −

Graph the parabola 2 1y x+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since 0 + 02 ≤ 1 is true, shade the same side of the parabola as (0, 0). Graph the parabola

2 1y x= − . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 20 0 1≥ − is true, shade the same side of the parabola as (0, 0). The overlapping region is the solution.

43.

0

0

2 6

2 6

x

y

x y

x y

≥≥

+ ≤+ ≤

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 6x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 6x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices:

The x-axis and y-axis intersect at (0, 0). The intersection of 2 6x y+ = and the y-axis is (0, 3). The intersection of 2 6x y+ = and the x-axis is (3, 0). To find the intersection of 2 6x y+ = and 2 6x y+ = , solve the system:

2 6

2 6

x y

x y

+ =+ =

Solve the first equation for x: 6 2x y= − . Substitute and solve: 2(6 2 ) 6

12 4 6

12 3 6

3 6

2

y y

y y

y

y

y

− + =− + =

− =− = −

=

6 2(2) 2x = − = The point of intersection is (2, 2). The four corner points are (0, 0), (0, 3), (3, 0), and (2, 2).


1147


44.

0

0

4

2 3 6

x

y

x y

x y

≥≥

+ ≥+ ≥

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 4x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 6x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded.

Find the vertices: The intersection of 4x y+ = and the y-axis is (0, 4). The intersection of 4x y+ = and the x-axis is (4, 0). The two corner points are (0, 4), and (4, 0).

=

45.

0

0

2

2 4

x

y

x y

x y

≥≥

+ ≥+ ≥

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 + 0 2 is false, shade the opposite side of the line from (0, 0). Graph the line 2 4x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded.

Find the vertices: The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 2 4x y+ = and the y-axis is (0, 4). The two corner points are (2, 0), and (0, 4).

46.

0

0

3 6

2 2

x

y

x y

x y

≥≥

+ ≤+ ≤

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 3 6x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of 0 and 0x y= = is (0, 0). The intersection of 2 2x y+ = and the x-axis is (1, 0). The intersection of 2 2x y+ = and the y-axis is (0, 2). The three corner points are (0, 0), (1, 0), and (0, 2).



47.

0

0

2

2 3 12

3 12

x

y

x y

x y

x y

≥≥

+ ≥+ ≤+ ≤

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 12x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 12+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 3 12x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 3(0) 0 12+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 2 3 12x y+ = and the y-axis is (0, 4). The intersection of 3 12x y+ = and the x-axis is (4, 0).

To find the intersection of 2 3 12x y+ = and 3 12x y+ = , solve the system:

2 3 12

3 12

x y

x y

+ =+ =

Solve the second equation for y: 12 3y x= − . Substitute and solve: 2 3(12 3 ) 12

2 36 9 12

7 24

247

x x

x x

x

x

+ − =+ − =

− = −

=

24 72 1212 3 12

7 7 7y = − = − =

The point of intersection is 24 12

,7 7

.

The five corner points are (0, 2), (0, 4), (2, 0),

(4, 0), and 24 12

,7 7

.

48.

0

0

1

7

2 10

x

y

x y

x y

x y

≥≥

+ ≥+ ≤+ ≤

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 1x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 1+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 7x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 7+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of 1x y+ = and the y-axis is (0, 1). The intersection of 1x y+ = and the x-axis is (1, 0).The intersection of 2 10x y+ = and the y-axis is (0, 10). The intersection of 2 10x y+ = and the x-axis is (0, 5).To find the intersection of 2 10 and 7x y x y+ = + = , solve the system:

2 10

7

x y

x y

+ =+ =

Solve the second equation for y: 7y x= − . Substitute and solve: 2 7 10 3x x x+ − = =

7 3 4y = − = The point of intersection is (3, 4). The five corner points are (0, 7), (5, 0), (0, 1), (1, 0) and (3, 4).


1149


49.

0

0

2

8

2 10

x

y

x y

x y

x y

≥≥

+ ≥+ ≤+ ≤

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 8x y+ = and the y-axis is (0, 8). The intersection of 2 10x y+ = and the x-axis is (5, 0). To find the intersection of

8 and 2 10x y x y+ = + = , solve the system:

8

2 10

x y

x y

+ =+ =

Solve the first equation for y: 8y x= − . Substitute and solve: 2 8 10

2

x x

x

+ − ==

8 2 6y = − = The point of intersection is (2, 6). The five corner points are (0, 2), (0, 8), (2, 0), (5, 0), and (2, 6).

50.

0

0

2

8

2 1

x

y

x y

x y

x y

≥≥

+ ≥+ ≤+ ≥

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 1x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 8x y+ = and the y-axis is (0, 8). The intersection of 8x y+ = and the x-axis is (8, 0). The four corner points are (0, 2), (0, 8), (2, 0), and (8, 0).



51.

0

0

2 1

2 10

x

y

x y

x y

≥≥

+ ≥+ ≤

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 1x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 10+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of 2 1x y+ = and the y-axis is (0, 0.5). The intersection of 2 1x y+ = and the x-axis is (1, 0). The intersection of 2 10x y+ = and the y-axis is (0, 5). The intersection of

2 10x y+ = and the x-axis is (10, 0). The four corner points are (0, 0.5), (0, 5), (1, 0), and (10, 0).

52.

0

0

2 1

2 10

2

8

x

y

x y

x y

x y

x y

≥≥

+ ≥+ ≤

+ ≥+ ≤

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 1x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses . Choose a

test point not on the line, such as (0, 0). Since 0 2(0) 10+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 2 10x y+ = and the y-axis is (0, 5). The intersection of 8x y+ = and the x-axis is (8, 0). To find the intersection of

8x y+ = and 2 10x y+ = , solve the system:

8

2 10

x y

x y

+ =+ =

Solve the first equation for x: 8x y= − . Substitute and solve: (8 ) 2 10

2

y y

y

− + ==

8 2 6x = − = The point of intersection is (6, 2). The five corner points are (0, 2), (0, 5), (2, 0), (8, 0), and (6, 2).

53. The system of linear inequalities is: 4

6

0

0

x

x y

x

y

≤+ ≤

≥≥


1151



2

6

0

0

y

x y

x

x

y

≤+ ≥

≤≥≥


15

50

0

0

x

y

x y

x y

x

≤≥

+ ≤− ≤

≥


5

3 4 12

2 8

0

y

x

x y

x y

x

≤≤

+ ≥− ≤

≥

57. a. Let x = the amount invested in Treasury bills, and let y = the amount invested in corporate bonds. The constraints are:

50,000x y+ ≤ because the total investment cannot exceed $50,000.

35,000x ≥ because the amount invested in Treasury bills must be at least $35,000.

10,000y ≤ because the amount invested in corporate bonds must not exceed $10,000.

0, 0x y≥ ≥ because a non-negative amount must be invested. The system is

50,000

35,000

10,000

0

0

x y

x

y

x

y

+ ≤≥≤≥≥

b. Graph the system.

The corner points are (35,000, 0),

(35,000, 10,000), (40,000, 10,000), (50,000, 0).

58. a. Let x = the # of standard model trucks, and let y = the # of deluxe model trucks. The constraints are:

0, 0x y≥ ≥ because a non-negative number of trucks must be manufactured. 2 3 80x y+ ≤ because the total painting hours worked cannot exceed 80. 3 4 120x y+ ≤ because the total detailing hours worked cannot exceed 120. The system is

0

0

2 3 80

3 4 120

x

y

x y

x y

≥≥+ ≤+ ≤


The corner points are ( ) ( )800,0 , 0. , 40,0

3.



59. a. Let x = the # of packages of the economy blend, and let y = the # of packages of the superior blend. The constraints are:

0, 0x y≥ ≥ because a non-negative # of packages must be produced. 4 8 75 16x y+ ≤ ⋅ because the total amount of “A grade” coffee cannot exceed 75 pounds. (Note: 75 pounds = (75)(16) ounces.) 12 8 120 16x y+ ≤ ⋅ because the total amount of “B grade” coffee cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) Simplifying the inequalities, we obtain:

4 8 75 16

2 75 4

2 300

x y

x y

x y

+ ≤ ⋅+ ≤ ⋅+ ≤

12 8 120 16

3 2 120 4

3 2 480

x y

x y

x y

+ ≤ ⋅+ ≤ ⋅+ ≤

The system is: 0

0

2 300

3 2 480

x

y

x y

x y

≥≥+ ≤+ ≤


The corner points are (0, 0), (0, 150), (90, 105), (160, 0).

60. a. Let x = the # of lower-priced packages, and let y = the # of quality packages. The constraints are:

0, 0x y≥ ≥ because a non-negative # of packages must be produced. 8 6 120 16x y+ ≤ ⋅ because the total amount of peanuts cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) 4 6 90 16x y+ ≤ ⋅ because the total amount of cashews cannot exceed 90 pounds. (Note: 90 pounds = (90)(16) ounces.) Simplifying the inequalities, we obtain:

8 6 120 164 3 120 84 3 960

x yx yx y

+ ≤ ⋅+ ≤ ⋅+ ≤

4 6 90 162 3 90 82 3 720

x yx yx y

+ ≤ ⋅+ ≤ ⋅+ ≤

The system is 4 3 9602 3 720

0; 0

x yx y

x y

+ ≤+ ≤

≥ ≥



61. a. Let x = the # of microwaves, and let y = the # of printers. The constraints are:

0, 0x y≥ ≥ because a non-negative # of items must be shipped. 30 20 1600x y+ ≤ because a total cargo weight cannot exceed 1600 pounds. 2 3 150x y+ ≤ because the total cargo volume cannot exceed 150 cubic feet. Note that the inequality 30 20 1600x y+ ≤ can be simplified: 3 2 160x y+ ≤ . The system is:

3 2 1602 3 150

0; 0

x yx y

x y

+ ≤+ ≤

≥ ≥


The corner points are (0, 0), (0, 50), (36, 26),

1603

,0 .

Section 11.8: Linear Programming

1153


Section 11.8

1. objective function

2. True

3. z x y= +

Vertex Value of

(0, 3) 0 3 3

(0, 6) 0 6 6

(5, 6) 5 6 11

(5, 2) 5 2 7

(4, 0) 4 0 4

z x y

z

z

z

z

z

= += + == + == + == + == + =

The maximum value is 11 at (5, 6), and the minimum value is 3 at (0, 3).

4. 2 3z x y= +

Vertex Value of 2 3

(0, 3) 2(0) 3(3) 9

(0, 6) 2(0) 3(6) 18

(5, 6) 2(5) 3(6) 28

(5, 2) 2(5) 3(2) 16

(4, 0) 2(4) 3(0) 8

z x y

z

z

z

z

z

= += + == + == + == + == + =


5. 10z x y= +

Vertex Value of 10

(0, 3) 0 10(3) 30

(0, 6) 0 10(6) 60

(5, 6) 5 10(6) 65

(5, 2) 5 10(2) 25

(4, 0) 4 10(0) 4

z x y

z

z

z

z

z

= += + == + == + == + == + =


6. 10z x y= +

Vertex Value of 10

(0, 3) 10(0) 3 3

(0, 6) 10(0) 6 6

(5, 6) 10(5) 6 56

(5, 2) 10(5) 2 52

(4, 0) 10(4) 0 40

z x y

z

z

z

z

z

= += + == + == + == + == + =


7. 5 7z x y= +

Vertex Value of 5 7

(0, 3) 5(0) 7(3) 21

(0, 6) 5(0) 7(6) 42

(5, 6) 5(5) 7(6) 67

(5, 2) 5(5) 7(2) 39

(4, 0) 5(4) 7(0) 20

z x y

z

z

z

z

z

= += + == + == + == + == + =


8. 7 5z x y= +

Vertex Value of 7 5

(0, 3) 7(0) 5(3) 15

(0, 6) 7(0) 5(6) 30

(5, 6) 7(5) 5(6) 65

(5, 2) 7(5) 5(2) 45

(4, 0) 7(4) 5(0) 28

z x y

z

z

z

z

z

= += + == + == + == + == + =


9. Maximize 2z x y= + subject to 0,x ≥ 0, 6,y x y≥ + ≤ 1x y+ ≥ . Graph the

constraints.

y

x(6,0)

(0,6)

(0,1)

(1,0)

x + y = 6

x + y = 1 The corner points are (0, 1), (1, 0), (0, 6), (6, 0).

Evaluate the objective function:

Vertex Value of 2

(0, 1) 2(0) 1 1

(0, 6) 2(0) 6 6

(1, 0) 2(1) 0 2

(6, 0) 2(6) 0 12

z x y

z

z

z

z

= += + == + == + =

= + =

The maximum value is 12 at (6, 0).



10. Maximize 3z x y= + subject to 0,x ≥ 0y ≥ , 3x y+ ≥ 5, 7x y≤ ≤ . Graph the constraints.

y

x

x = 5

y = 7

x + y = 3

(5,7)(0,7)

(5,0)

(0,3)

(3,0)

The corner points are (0, 3), (3, 0), (0, 7), (5, 0),

(5, 7). Evaluate the objective function:

Vertex Value of 3

(0, 3) 0 3(3) 9

(3, 0) 3 3(0) 3

(0, 7) 0 3(7) 21

(5, 0) 5 3(0) 5

(5, 7) 5 3(7) 26

z x y

z

z

z

z

z

= += + == + =

= + == + =

= + =


11. Minimize 2 5z x y= + subject to 0,x ≥ 0, 2, 5,y x y x≥ + ≥ ≤ 3y ≤ . Graph the

constraints.

y

x

(5,3)

(5,0)

(0,3)

(2,0)

(0,2)

x = 5

y = 3

x + y = 2 The corner points are (0, 2), (2, 0), (0, 3), (5, 0),


Vertex Value of 2 5

(0, 2) 2(0) 5(2) 10

(0, 3) 2(0) 5(3) 15

(2, 0) 2(2) 5(0) 4

(5, 0) 2(5) 5(0) 10

(5, 3) 2(5) 5(3) 25

z x y

z

z

z

z

z

= += + == + == + == + == + =

The minimum value is 4 at (2, 0).

12. Minimize 3 4z x y= + subject to 0x ≥ , 0, 2 3 6,y x y≥ + ≥ 8x y+ ≤ . Graph the

constraints.

y

x

x + y = 8

2x + 3y = 6

(0,8)

(0,2)

(3,0) (8,0)



Vertex Value of 3 4

(0, 2) 3(0) 4(2) 8

(3, 0) 3(3) 4(0) 9

(0, 8) 3(0) 4(8) 32

(8, 0) 3(8) 4(0) 24

z x y

z

z

z

z

= += + == + =

= + == + =


13. Maximize 3 5z x y= + subject to 0,x ≥ 0,y ≥ 2,x y+ ≥ 2 3 12,x y+ ≤ 3 2 12x y+ ≤ . Graph

the constraints.

y

x

(2.4,2.4)

(0,4)

(0,2)

(2,0) (4,0)

3x + 2y = 12

2x + 3y = 12

x + y = 2 To find the intersection of 2 3 12x y+ = and

3 2 12x y+ = , solve the system:

2 3 123 2 12

x yx y

+ =+ =

Solve the second equation for y: 3

62

y x= −

Substitute and solve:


1155


32 3 6 12

29

2 18 1225

62

125

x x

x x

x

x

+ − =

+ − =

− = −

=

3 12 18 126 6

2 5 5 5y = − = − =

The point of intersection is ( )2.4, 2.4 .

The corner points are (0, 2), (2, 0), (0, 4), (4, 0), (2.4, 2.4). Evaluate the objective function:

Vertex Value of 3 5

(0, 2) 3(0) 5(2) 10

(0, 4) 3(0) 5(4) 20

(2, 0) 3(2) 5(0) 6

(4, 0) 3(4) 5(0) 12

(2.4, 2.4) 3(2.4) 5(2.4) 19.2

z x y

z

z

z

z

z

= += + == + == + == + =

= + =


14. Maximize 5 3z x y= + subject to 0,x ≥ 0, 2, 8, 2 10y x y x y x y≥ + ≥ + ≤ + ≤ . Graph

the constraints.

y

x

x + y = 8

2x + y = 10

x + y = 2

(2,0)

(0,2)

(0,8)

(5,0)

(2,6)

To find the intersection of 8x y+ = and

2 10x y+ = , solve the system:

8

2 10

x y

x y

+ =+ =

Solve the first equation for y: 8y x= − . Substitute and solve: 2 8 10

2

x x

x

+ − ==

8 2 6y = − = The point of intersection is (2, 6). The corner points are (0, 2), (2, 0), (0, 8), (5, 0), (2, 6). Evaluate the objective function:

Vertex Value of 5 3

(0, 2) 5(0) 3(2) 6

(0, 8) 5(0) 3(8) 24

(2, 0) 5(2) 3(0) 10

(5, 0) 5(5) 3(0) 25

(2, 6) 5(2) 3(6) 28

z x y

z

z

z

z

z

= += + == + == + == + == + =


15. Minimize 5 4z x y= + subject to 0,x ≥ 0, 2, 2 3 12, 3 12y x y x y x y≥ + ≥ + ≤ + ≤ . Graph

the constraints.

y

x

(0,4)

(0,2)

(2,0) (4,0)

x + y = 2

2x + 3y = 12

3x + y = 12

24

7, 12

7( )

To find the intersection of 2 3 12x y+ = and


2 3 12

3 12

x y

x y

+ =+ =

Solve the second equation for y: 12 3y x= − Substitute and solve:

247

2 3(12 3 ) 122 36 9 12

7 24

x xx x

x

x

+ − =+ − =

− = −=

( ) 7224 127 7 712 3 12y = − = − =

The point of intersection is 24 12,

7 7.

The corner points are (0, 2), (2, 0), (0, 4), (4, 0), 24 12

,7 7

. Evaluate the objective function:

( ) ( ) ( )24 12 24 127 7 7 7

Vertex Value of 5 4

(0, 2) 5(0) 4(2) 8

(0, 4) 5(0) 4(4) 16

(2, 0) 5(2) 4(0) 10

(4, 0) 5(4) 4(0) 20

, 5 4 24

z x y

z

z

z

z

z

= += + == + == + == + =

= + =




16. Minimize 2 3z x y= + subject to 0,x ≥ 0, 3, 9, 3 6y x y x y x y≥ + ≥ + ≤ + ≥ . Graph

the constraints.

y

x

x + y = 9

x + y = 3x + 3y = 6

(0,9)

(0,3)

(6,0)(9,0)

32 , 3

2( )

To find the intersection of 3x y+ = and 3 6x y+ = , solve the system:

3

3 6

x y

x y

+ =+ =

Solve the first equation for y: 3y x= − . Substitute and solve:

3(3 ) 6

9 3 6

2 3

32

x x

x x

x

x

+ − =+ − =

− = −

=

3 33

2 2y = − =


,2 2

.


,2 2

. Evaluate the objective function:

Vertex Value of 2 3

(0, 3) 2(0) 3(3) 9

(0, 9) 2(0) 3(9) 27

(6, 0) 2(6) 3(0) 12

(9, 0) 2(9) 3(0) 18

3 3 3 3 15, 2 4

2 2 2 2 2

z x y

z

z

z

z

z

= += + =

= + == + == + =

= + =

The minimum value is 152

at 3 3

,2 2

.

17. Maximize 5 2z x y= + subject to 0,x ≥ 0, 10, 2 10, 2 10y x y x y x y≥ + ≤ + ≥ + ≥ .

Graph the constraints.

y

x

(0,10)

(10,0)

x + y = 10

2x + y = 10x + 2y = 10

103 , 10

3( )

To find the intersection of 2 10x y+ = and 2 10x y+ = , solve the system:

2 10

2 10

x y

x y

+ =+ =

Solve the first equation for y: 10 2y x= − . Substitute and solve:

2(10 2 ) 10

20 4 10

3 10

103

x x

x x

x

x

+ − =+ − =

− = −

=

10 20 1010 2 10

3 3 3y = − = − =

The point of intersection is (10/3 10/3). The corner points are (0, 10), (10, 0), (10/3, 10/3). Evaluate the objective function:

13

Vertex Value of 5 2

(0, 10) 5(0) 2(10) 20

(10, 0) 5(10) 2(0) 50

10 10 10 10 70, 5 2 23

3 3 3 3 3

z x y

z

z

z

= += + == + =

= + = =



1157


18. Maximize 2 4z x y= + subject to 0, 0, 2 4, 9x y x y x y≥ ≥ + ≥ + ≤ .


y

x

x + y = 9

2x + y = 4

(0,9)

(9,0)(2,0)

(0,4)



Vertex Value of 2 4

(0, 9) 2(0) 4(9) 36

(9, 0) 2(9) 4(0) 18

(0, 4) 2(0) 4(4) 16

(2, 0) 2(2) 4(0) 4

z x y

z

z

z

z

= += + == + == + == + =


19. Let x = the number of downhill skis produced, and let y = the number of cross-country skis produced. The total profit is: 70 50P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A positive number of skis must be produced.

2 40x y+ ≤ Manufacturing time available. 32x y+ ≤ Finishing time available. Graph the constraints.

y

x

(8,24)

(20,0)(0,0)

(0,32)

2x + y = 40

x + y = 32



32

2 40

x y

x y

+ =+ =

Solve the first equation for y: 32y x= − . Substitute and solve: 2 (32 ) 40

8

x x

x

+ − ==

32 8 24y = − = The point of intersection is (8, 24). The corner points are (0, 0), (0, 32), (20, 0), (8, 24). Evaluate the objective function:

Vertex Value of 70 50

(0, 0) 70(0) 50(0) 0

(0, 32) 70(0) 50(32) 1600

(20, 0) 70(20) 50(0) 1400

(8, 24) 70(8) 50(24) 1760

P x y

P

P

P

P

= += + =

= + == + == + =

The maximum profit is $1760, when 8 downhill skis and 24 cross-country skis are produced.

With the increase of the manufacturing time to 48 hours, we do the following: The constraints are:

0, 0x y≥ ≥ A positive number of skis must be produced.

2 48x y+ ≤ Manufacturing time available. 32x y+ ≤ Finishing time available. Graph the constraints.

y

x

(16,16)

(0,32)

(24,0)(0,0)

2x + y = 48

x + y = 32



32

2 48

x y

x y

+ =+ =

Solve the first equation for y: 32y x= − . Substitute and solve: 2 (32 ) 48

16

x x

x

+ − ==

32 16 16y = − = The point of intersection is (16, 16). The corner points are (0, 0), (0, 32), (24, 0), (16, 16). Evaluate the objective function:




(0, 0) 70(0) 50(0) 0

(0, 32) 70(0) 50(32) 1600

(24, 0) 70(24) 50(0) 1680

(16, 16) 70(16) 50(16) 1920

P x y

P

P

P

P

= += + =

= + == + == + =

The maximum profit is $1920, when 16 downhill skis and 16 cross-country skis are produced.

20. Let x = the number of acres of soybeans planted , and let y = the number of acres of wheat planted. The total profit is: 180 100P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A non-negative number of acres must be planted.

70x y+ ≤ Acres available to plant. 60 30 1800x y+ ≤ Money available for

preparation. 3 4 120x y+ ≤ Workdays available. Graph the constraints.

y

x

x + y = 70

60x + 30y = 1800

3x + 4y = 120

(0,30)

(30,0)(0,0)

(24,12)


3 4 120x y+ = , solve the system:

60 30 1800

3 4 120

x y

x y

+ =+ =

Solve the first equation for y: 60 30 1800

2 60

60 2

x y

x y

y x

+ =+ =

= −

Substitute and solve: 3 4(60 2 ) 120

3 240 8 120

5 120

24

x x

x x

x

x

+ − =+ − =

− = −=

60 2(24) 12y = − = The point of intersection is (24, 12). The corner points are (0, 0), (0, 30), (30, 0), (24, 12). Evaluate the objective function:


(0, 0) 180(0) 100(0) 0

(0, 30) 180(0) 100(30) 3000

(30, 0) 180(30) 100(0) 5400

(24, 12) 180(24) 100(12) 5520

P x y

P

P

P

P

= += + =

= + == + == + =

The maximum profit is $5520, when 24 acres of soybeans and 12 acres of wheat are planted.

With the increase of the preparation costs to $2400, we do the following: The constraints are:

0, 0x y≥ ≥ A non-negative number of acres must be planted.

70x y+ ≤ Acres available to plant. 60 30 2400x y+ ≤ Money available for

preparation. 3 4 120x y+ ≤ Workdays available. Graph the constraints.

y

x

x + y = 70

3x + 4y = 120

60x + 30y = 2400

(0,30)

(40,0)(0,0)

The corner points are (0, 0), (0, 30), (40, 0).



(0, 0) 180(0) 100(0) 0

(0, 30) 180(0) 100(30) 3000

(40, 0) 180(40) 100(0) 7200

P x y

P

P

P

= += + =

= + == + =

The maximum profit is $7200, when 40 acres of soybeans and 0 acres of wheat are planted.

21. Let x = the number of rectangular tables rented, and let y = the number of round tables rented. The cost for the tables is: 28 52C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A non-negative number of tables must be used.

35x y+ ≤ Maximum number of tables. 6 10 250x y+ ≥ Number of guests. 15x ≤ Rectangular tables available.


1159



y

x

6 + 10 = 250x y

x y + = 35

(15, 20)

x = 15

(0, 25)

(0, 35)

(15, 16)

25

10




(0, 25) 28(0) 52(25) 1300

(0, 35) 28(0) 52(35) 1820

(15, 20) 28(15) 52(20) 1460

(15,16) 28(15) 52(16) 1252

C x y

C

C

C

C

= += + == + == + == + =

Kathleen should rent 15 rectangular tables and 16 round tables in order to minimize the cost. The minimum cost is $1252.00.

22. Let x = the number of buses rented, and let y = the number of vans rented. The cost for the vehicles is: 975 350C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A non-negative number of buses and vans must be used.

40 8 320x y+ ≥ Number of regular seats. 3 36x y+ ≥ Number of handicapped seats. Graph the constraints.

y

xx y + 3 = 36

40 + 8 = 320x y

(6, 10)(0, 12)

(0, 40)

(36, 0)

25

25



40 8 320

3 36

x y

x y

+ =+ =

Solve the second equation for x:

3 36

3 36

x y

x y

+ == − +

Substitute and solve: 40( 3 36) 8 320

120 1440 8 320

112 1120

10

y y

y y

y

y

− + + =− + + =

− = −=

3(10) 36 30 36 6x = − + = − + = The point of intersection is (6, 10). The corner points are (0, 40), (6, 10), and (36, 0). Evaluate the objective function:


(0, 40) 975(0) 350(40) 14,000

(6,10) 975(6) 350(10) 9350

(36, 0) 975(36) 350(0) 35,100

C x y

C

C

C

= += + == + == + =

The college should rent 6 buses and 10 vans for a minimum cost of $9350.00.

23. Let x = the amount invested in junk bonds, and let y = the amount invested in Treasury bills. The total income is: 0.09 0.07I x y= + . Income is to be maximized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A non-negative amount must be invested.

20,000x y+ ≤ Total investment cannot exceed $20,000.

12,000x ≤ Amount invested in junk bonds must not exceed $12,000.

8000y ≥ Amount invested in Treasury bills must be at least $8000.

a. y x≥ Amount invested in Treasury bills must be equal to or greater than the amount invested in junk bonds.


x + y = 20000

y = x

x = 12000

y = 8000

(0,20000)

(0,8000) (8000,8000)

(10000,10000)



The corner points are (0, 20,000), (0, 8000), (8000, 8000), (10,000, 10,000). Evaluate the objective function:

Vertex Value of 0.09 0.07

(0, 20000) 0.09(0) 0.07(20000)

1400

(0, 8000) 0.09(0) 0.07(8000)

560

(8000, 8000) 0.09(8000) 0.07(8000)

1280

(10000, 10000) 0.09(10000) 0.07(10000)

1600

I x y

I

I

I

I

= += +== +== +== +=

The maximum income is $1600, when $10,000 is invested in junk bonds and $10,000 is invested in Treasury bills.

b. y x≤ Amount invested in Treasury bills must not exceed the amount invested in junk bonds.


x + y = 20000

y = x

x = 12000

y = 8000(8000,8000)

(10000,10000)

(12000,8000)

The corner points are (12,000, 8000),

(8000, 8000), (10,000, 10,000). Evaluate the objective function:


(12000, 8000) 0.09(12000) 0.07(8000)

1640

(8000, 8000) 0.09(8000) 0.07(8000)

1280

(10000, 10000) 0.09(10000) 0.07(10000)

1600

I x y

I

I

I

= += +== +=

= +=

The maximum income is $1640, when $12,000 is invested in junk bonds and $8000 is invested in Treasury bills.

24. Let x = the number of hours that machine 1 is operated, and let y = the number of hours that machine 2 is operated. The total cost is:

50 30C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A positive number of hours must be used.

10x ≤ Time used on machine 1. 10y ≤ Time used on machine 2. 60 40 240x y+ ≥ 8-inch pliers to be produced. 70 20 140x y+ ≥ 6-inch pliers to be produced. Graph the constraints.

(10,10)(0,10)

(10,0)(4,0)

(0,7)

60x + 40y = 24070x + 20y = 140

12 , 21

4( )

To find the intersection of 60 40 240x y+ = and 70 20 140x y+ = , solve the system:

60 40 240

70 20 140

x y

x y

+ =+ =

Divide the first equation by 2− and add the result to the second equation:

30 20 120

70 20 140

x y

x y

− − = −+ =

40 20

20 140 2

x

x

=

= =

Substitute and solve: 1

60 40 240230 40 240

40 210

210 21 15

40 4 4

y

y

y

y

+ =

+ ==

= = =


, 52 4

.


1161



(10, 0), (10, 10), 1 1

, 52 4

. Evaluate the

objective function: Vertex Value of 50 30

(0, 7) 50(0) 30(7) 210

(0, 10) 50(0) 30(10) 300

(4, 0) 50(4) 30(0) 200

(10, 0) 50(10) 30(0) 500

(10, 10) 50(10) 30(10) 800

1 1 1 1, 5 50 30 5 182.50

2 4 2 4

C x y

C

C

C

C

C

P

= += + == + == + == + == + =

= + =

The minimum cost is $182.50, when machine 1

is used for 12

hour and machine 2 is used for 1

54

hours.

25. Let x = the number of pounds of ground beef, and let y = the number of pounds of ground pork. The total cost is: 0.75 0.45C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A positive number of pounds must be used.

200x ≤ Only 200 pounds of ground beef are available.

50y ≥ At least 50 pounds of ground pork must be used.

0.75 0.60 0.70( )x y x y+ ≥ + Leanness condition

(Note that the last equation will simplify to 12

y x≤ .) Graph the constraints.

y

x

(200,100)

(200,50)(100,50)

The corner points are (100, 50), (200, 50),



(100, 50) 0.75(100) 0.45(50) 97.50

(200, 50) 0.75(200) 0.45(50) 172.50

(200, 100) 0.75(200) 0.45(100) 195

C x y

C

C

C

= += + == + == + =

The minimum cost is $97.50, when 100 pounds of ground beef and 50 pounds of ground pork are used.

26. Let x = the number of gallons of regular, and let y = the number of gallons of premium. The

total profit is: 0.75 0.90P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A positive number of gallons must be used.

14

y x≥ At least one gallon of premium

for every 4 gallons of regular. 5 6 3000x y+ ≤ Daily shipping weight limit.

24 20 16(725)x y+ ≤ Available flavoring.

12 20 16(425)x y+ ≤ Available milk-fat (Note: the last two inequalities simplify to

6 5 2900x y+ ≤ and 3 5 1700x y+ ≤ .) Graph the constraints.

y

x

5 + 6 = 3000x y

(0, 340) (400, 100)

300

15014

y x=

24 + 20 = 16(725)x y

12 + 20 = 16(425)x y

The corner points are (0, 0), (400, 100), (0, 340). Evaluate the objective function:


(0, 0) 0.75(0) 0.90(0) 0

(400,100) 0.75(400) 0.90(100) 390

(0, 340) 0.75(0) 0.90(340) 306

P x y

P

P

P

= += + == + == + =

Mom and Pop should produce 400 gallons of regular and 100 gallons of premium ice cream. The maximum profit is $390.00.



27. Let x = the number of racing skates manufactured, and let y = the number of figure skates manufactured. The total profit is:

10 12P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A positive number of skates must be manufactured.

6 4 120x y+ ≤ Only 120 hours are available for fabrication.

2 40x y+ ≤ Only 40 hours are available for finishing.


y

x

(10,15)

(20,0)(0,0)

(0,20)


+2 40x y = , solve the system:

6 4 120

2 40

x y

x y

+ =+ =

Solve the second equation for x: 40 2x y= − Substitute and solve: 6(40 2 ) 4 120

240 12 4 120

8 120

15

y y

y y

y

y

− + =− + =

− = −=

40 2(15) 10x = − = The point of intersection is (10, 15). The corner points are (0, 0), (0, 20), (20, 0), (10, 15). Evaluate the objective function:


(0, 0) 10(0) 12(0) 0

(0, 20) 10(0) 12(20) 240

(20, 0) 10(20) 12(0) 200

(10, 15) 10(10) 12(15) 280

P x y

P

P

P

P

= += + =

= + == + == + =

The maximum profit is $280, when 10 racing skates and 15 figure skates are produced.

28. Let x = the amount placed in the AAA bond. Let y = the amount placed in a CD. The total return is: 0.08 0.04R x y= + . Return is to be maximized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A positive amount must be invested in each.

50,000x y+ ≤ Total investment cannot exceed $50,000.

20,000x ≤ Investment in the AAA bond cannot exceed $20,000.

15,000y ≥ Investment in the CD must be at least $15,000.

y x≥ Investment in the CD must exceed or equal the investment in the bond.


y

x

y = x

x + y = 50000

x = 20000

y = 15000

(0,50000)

(0,15000)(15000,15000)

(20000,30000)

(20000,20000)

The corner points are (0, 50,000), (0, 15,000),

(15,000, 15,000), (20,000, 20,000), (20,000, 30,000). Evaluate the objective function:


(0, 50000) 0.08(0) 0.04(50000)

2000

(0, 15000) 0.08(0) 0.04(15000)

600

(15000, 15000) 0.08(15000) 0.04(15000)

1800

(20000, 20000) 0.08(20000) 0.04(20000)

2400

(20000, 30000

R x y

R

R

R

R

= += +== +== +== +=

) 0.08(20000) 0.04(30000)

2800

R = +=

The maximum return is $2800, when $20,000 is invested in a AAA bond and $30,000 is invested in a CD.


1163


29. Let x = the number of metal fasteners, and let y = the number of plastic fasteners. The total cost is: 9 4C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:

2, 2x y≥ ≥ At least 2 of each fastener must be made.

6x y+ ≥ At least 6 fasteners are needed. 4 2 24x y+ ≤ Only 24 hours are available. Graph the constraints.

y

x

(2,8)

(5,2)(4,2)

(2,4)



Vertex Value of 9 4

(2, 4) 9(2) 4(4) 34

(2, 8) 9(2) 4(8) 50

(4, 2) 9(4) 4(2) 44

(5, 2) 9(5) 4(2) 53

C x y

C

C

C

C

= += + == + == + == + =

The minimum cost is $34, when 2 metal fasteners and 4 plastic fasteners are ordered.

30. Let x = the amount of “Gourmet Dog,” and let y = the amount of “Chow Hound.” The total

cost is: 0.40 0.32C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:

0, 0x y≥ ≥ A non-negative number of cans must be purchased.

20 35 1175x y+ ≥ At least 1175 units of vitamins per month.

75 50 2375x y+ ≥ At least 2375 calories per month.

60x y+ ≤ Storage space for 60 cans. Graph the constraints.

y

x

x + y = 60

75x+50y=2375

20x+35y=1175

(0,47.5)

(0,60)

(60,0)

(58.75,0)

(15,25)

The corner points are (0, 47.5), (0, 60), (60, 0),

(58.75, 0), (15, 25). Evaluate the objective function:


(0, 47.5) 0.40(0) 0.32(47.5) 15.20

(0, 60) 0.40(0) 0.32(60) 19.20

(60, 0) 0.40(60) 0.32(0) 24.00

(58.75, 0) 0.40(58.75) 0.32(0) 23.50

(15, 25) 0.40(15) 0.32(25) 14.00

C x y

C

C

C

C

C

= += + == + == + =

= + == + =

The minimum cost is $14, when 15 cans of "Gourmet Dog" and 25 cans of “Chow Hound” are purchased.

31. Let x = the number of first class seats, and let y = the number of coach seats. Using the hint,

the revenue from x first class seats and y coach seats is ,Fx Cy+ where 0.F C> > Thus, R Fx Cy= + is the objective function to be maximized. The constraints are: 8 16x≤ ≤ Restriction on first class seats. 80 120y≤ ≤ Restriction on coach seats.

a. 1

12xy

≤ Ratio of seats.

The constraints are: 8 16x≤ ≤ 80 120y≤ ≤ 12x y≤ Graph the constraints.



The corner points are (8, 96), (8, 120), and (10, 120). Evaluate the objective function:

Vertex Value of

(8, 96) 8 96

(8, 120) 8 120

(10, 120) 10 120

R Fx Cy

R F C

R F C

R F C

= += += += +

Since 0,C > 120 96 ,C C> so 8 120 8 96 .F C F C+ > + Since 0,F > 10 8 ,F F> so 10 120 8 120 .F C F C+ > + Thus, the maximum revenue occurs when the aircraft is configured with 10 first class seats and 120 coach seats.

b. 18

xy

≤

The constraints are: 8 16x≤ ≤ 80 120y≤ ≤ 8x y≤ Graph the constraints.

The corner points are (8, 80), (8, 120),

(15, 120), and (10, 80). Evaluate the objective function:

Vertex Value of

(8, 80) 8 80

(8, 120) 8 120

(15, 120) 15 120

(10, 80) 10 80

R Fx Cy

R F C

R F C

R F C

R F C

= += += += += +

Since 0F > and 0,C > 120 96 ,C C> the maximum value of R occurs at (15, 120). The maximum revenue occurs when the aircraft is configured with 15 first class seats and 120 coach seats.

c. Answers will vary.


Chapter 11 Review Exercises

1. 2 5

5 2 8

x y

x y

− =+ =

Solve the first equation for y: 2 5y x= − . Substitute and solve: 5 2(2 5) 8

5 4 10 8

9 18

2

x x

x x

x

x

+ − =+ − =

==

2(2) 5 4 5 1y = − = − = −


2. 3 4 4

13

2

x y

x y

− =

− =

Solve the second equation for x: 1

32

x y= +

Substitute into the first equation and solve: 1

3 3 4 423

9 4 42

55

212

y y

y y

y

y

+ − =

+ − =

=

=

1 13 2

2 2x = + =

The solution is 1

2,2

x y= = or 1

2,2

.

3. 2 4 0

3 2 4 0

x y

x y

− − =+ − =

Solve the first equation for x: 2 4x y= + Substitute into the second equation and solve: 3(2 4) 2 4 0

6 12 2 4 0

8 8

1

y y

y y

y

y

+ + − =+ + − =

= −= −

2( 1) 4 2x = − + =



1165


4. 2 5

3 4

y x

x y

= −= +

Substitute the first equation into the second equation and solve:

3(2 5) 4

6 15 4

5 11

115

x x

x x

x

x

= − += − +

− = −

=

11 32 5

5 5y = − = −

The solution is 11 3

,5 5

x y= = − or 11 3

,5 5

− .

5. 3 4 0

1 3 40

2 2 3

x y

x y

− + =

− + =

Multiply each side of the first equation by 3 and each side of the second equation by 6− and add:

3 9 12 0

3 9 8 0

4 0

x y

x y

− + =− + − =

=

There is no solution to the system. The system is inconsistent.

6. 2 3 13 0

3 2 0

x y

x y

+ − =− =

Multiply each side of the first equation by 2 and each side of the second equation by 3, and add to eliminate y:

4 6 26 0

9 6 0

13 26 0

13 26

2

x y

x y

x

x

x

+ − =− =

− ===

Substitute and solve for y: 3(2) 2 0

2 6

3

y

y

y

− =− = −

=

The solution is 2, 3x y= = or (2, 3).

7. 2 5 10

4 10 20

x y

x y

+ =+ =

Multiply each side of the first equation by –2 and add to eliminate x:

4 10 20

4 10 20

0 0

x y

x y

− − = −+ =

=

The system is dependent. 2 5 10

5 2 10

22

5

x y

y x

y x

+ == − +

= − +

The solution is 2

25

y x= − + , x is any real number

or 2

( , ) 2, is any real number5

x y y x x= − + .

8. 2 6

2 3 13

3 2 3 16

x y z

x y z

x y z

+ − =− + = −− + = −

Multiply each side of the first equation by –2 and add to the second equation to eliminate x;

2 4 2 12

2 3 13

5 5 25

5

x y z

x y z

y z

y z

− − + = −− + = −

− + = −− =

Multiply each side of the first equation by –3 and add to the third equation to eliminate x:

3 6 3 18

3 2 3 16

8 6 34

x y z

x y z

y z

− − + = −− + = −− + = −

Multiply each side of the first result by 8 and add to the second result to eliminate y:

8 8 40

8 6 34

2 6

3

y z

y z

z

z

− =− + = −

− == −

Substituting and solving for the other variables: ( 3) 5

2

y

y

− − ==

2(2) ( 3) 6

4 3 6

1

x

x

x

+ − − =+ + =

= −

The solution is 1, 2, 3x y z= − = = − or ( 1, 2, 3)− − .

9. 2 4 15

2 4 27

5 6 2 3

x y z

x y z

x y z

− + = −+ − =− − = −

Multiply the first equation by 1− and the second



equation by 2, and then add to eliminate x: 2 4 15

2 4 8 54

8 9 69

x y z

x y z

y z

− + − =+ − =

− =

Multiply the second equation by 5− and add to the third equation to eliminate x:

5 10 20 135

5 6 2 3

16 18 138

x y z

x y z

y z

− − + = −− − = −− + = −

Multiply both sides of the first result by 2 and add to the second result to eliminate y: 16 18 138

16 18 138

0 0

y z

y z

− =− + = −

=

The system is dependent. 16 18 138

18 138 16

9 698 8

y z

z y

y z

− + = −+ =

= +

Substituting into the second equation and solving for x:

9 692 4 27

8 89 69

4 274 4

7 394 4

x z z

x z z

x z

+ + − =

+ + − =

= +


x z= + , 9 698 8

y z= + , z is

any real number or 7 39

( , , ) ,4 4

x y z x z= +


8 8y z z= + .

10. 4 3 15

3 5 5

7 5 9 10

x y z

x y z

x y z

− + =− + − = −− − − =

Multiply the first equation by 3 and then add the second equation to eliminate x:

3 12 9 45

3 5 5

11 4 40

x y z

x y z

y z

− + =− + − = −

− + =

Multiply the first equation by 7 and add to the

third equation to eliminate x: 7 28 21 105

7 5 9 10

33 12 115

115 11 4

3

x y z

x y z

y z

y z

− + =− − − =

− + =

− + =

Multiply the first result by 1− and adding it to the second result:

11 4 40

11511 4

3

5 0

3

y z

y z

− = −

− + =

= −

The system has no solution. The system is inconsistent.

11. 3 2 8

4 1

x y

x y

+ =+ = −

12. 2 5 2

5 3 8

2 0

x y z

x z

x y

+ + = −− =

− =

13. 1 0 3 4

2 4 1 5

1 2 5 2

1 3 0 ( 4) 4 4

2 1 4 5 3 9

1 5 2 2 4 4

A C

−+ = +

−+ + − −

= + + =− + +

14. 1 0 6 1 6 0 6 0

6 6 2 4 6 2 6 4 12 24

1 2 6( 1) 6 2 6 12

A

⋅ ⋅= ⋅ = ⋅ ⋅ =

− − ⋅ −

15. 1 0

4 3 02 4

1 1 21 2

1(4) 0(1) 1( 3) 0(1) 1(0) 0( 2)

2(4) 4(1) 2( 3) 4(1) 2(0) 4( 2)

1(4) 2(1) 1( 3) 2(1) 1(0) 2( 2)

4 3 0

12 2 8

2 5 4

AB−

= ⋅−

−+ − + + −

= + − + + −− + − − + − + −

−= − −

− −


1167


16. 3 4

4 3 01 5

1 1 25 2

4(3) 3(1) 0(5) 4( 4) 3(5) 0(2)

1(3) 1(1) 2(5) 1( 4) 1(5) 2(2)

9 31

6 3

BC

−−

= ⋅−

− + − − +=

+ − − + −−

=− −

17. 4 6

1 3A =


( )

( )

( )

1 2

2 1 2

12 21 2 6

6 3

12

1 2 11 26 3

4 6 1 0

1 3 0 1

Interchange1 3 0 1

and 4 6 1 0

1 3 0 1 4

0 6 1 4

1 3 0 1

0 1

1 0 1 3

0 1

r r

R r r

R r

R r r

→

→ = − +− −

→ = −−

−→ = − +

−

Thus, 1

1 21 26 3

1A− −

=−

.

18. 1 3 3

1 2 1

1 1 2

A =−


( )

2 1 2

3 1 3

2 2

1 2 1

3 2 3

1 3 3 1 0 0

1 2 1 0 1 0

1 1 2 0 0 1

1 3 3 1 0 0

0 1 2 1 1 0

0 4 1 1 0 1

1 3 3 1 0 0

0 1 2 1 1 0

0 4 1 1 0 1

1 0 3 2 3 03

0 1 2 1 1 0 4

0 0 7 3 4 1

R r r

R r r

R r

R r r

R r r

−

= − +→ − − −

= − +− − −

→ − = −− − −

− −= − +

→ −= +

−

( )13 37

3 4 17 7 7

5 9 37 7 7

1 3 11 1 27 7 7

2 3 23 4 17 7 7

1 0 3 2 3 0

0 1 2 1 1 0

0 0 1

1 0 03

0 1 0 2

0 0 1

R r

R r r

R r r

− −→ − =

−

−= +

→ −= − +

−

Thus,

5 9 37 7 7

1 1 1 27 7 73 4 17 7 7

A−

−= −

−.

19. 4 8

1 2A

−=

−


1 2

4 8 1 0

1 2 0 1

Interchange1 2 0 1

and 4 8 1 0 r r

−−

−→

−

( )

( )

2 1 2

1 1

1 2 0 1 4

0 0 1 4

1 2 0 1

0 0 1 4

R r r

R r

−→ = +

− −→ = −

There is no inverse because there is no way to obtain the identity on the left side. The matrix is singular.



20. 3 2 1

10 10 5

x y

x y

− =+ =


( )

( )

( )

2 1 2

1 2

2 1 2

12 2501

10

3 2 1

10 10 5

23 1 3161 2

Interchange161 2 and 23 1

161 2 30 50 5

161 2

0 1

R r r

r r

R r r

R r

−

−→ = − +

→−

→ = − +− −

→ = −

( )25

1 2 11

10

01 160 1

R r r→ = − +

The solution is 2 1

,5 10

x y= = or 2 1

,5 10

.

21. 5 6 3 6

4 7 2 3

3 7 1

x y z

x y z

x y z

− − =− − = −+ − =


( )1 2 1

2 1 2

3 1 3

1392 2 211

11 111

3 32

5 6 3 6

27 34

3 71 1

91 1 1

27 34

3 71 1

91 1 14

0 39 11 23

4 260 2

91 1 1

0 1

0 131 2

R r r

R r r

R r r

R r

R r

− −−− −−

−−− −→ = − +−

−= − +

→ −−= − +

− −−

−= −−→= −

9 6011 11

1 2 139211 11

3 2 31042411 11

1 0

0 1

0 0

R r r

R r r

−= − +

→ −= − +

( )9 6011 11

392 113 311 11 24

133

91 3 11113

3 22 3 213 11

3

1 0

0 1

0 0 1

1 0 0 9

0 1 0

0 0 1

R r

R r r

R r r

−→ − =

= +→

= +


9, ,3 3

x y z= = = or

13 139, ,

3 3.

22. 2 5

4 3 1

8 5

x y z

x y z

x y z

+ + =− − =+ − =


52 1 1

34 1 1

8 51 1

−−−

2 1 2

3 1 3

51 112 2 2

1 1253 1

2 23

13 1

1 2 15 23

3 2 3

52 1 12

0 3 5 9 4

0 3 5 15

1

0 1 3

0 3 5 15

1 0 1

0 1 3 3

0 0 0 6

R r r

R r r

R r

R r

R r r

R r r

= − +→ − − −

= − +− − −

=→

= −− − −

−= − +

→= +

−

There is no solution; the system is inconsistent.

23. 2 1

2 3 3

4 3 4 3

x z

x y

x y z

− =+ = −− − =


2 1 2

3 1 3

1 0 2 1

2 3 0 3

4 3 4 3

1 0 2 12

0 3 4 5 4

0 3 4 1

R r r

R r r

−−

− −−

= − +→ −

= − +− −


1169


( )54 12 23 3 3

1 0 2 1

0 1

0 3 4 1

R r

−→ − =

− −

( )543 2 33 3

1 0 2 1

0 1 3

0 0 8 6

R r r

−→ − = +

−

( )54 13 33 3 8

34

12

1 3 123 4

2 3 2334

1 0 2 1

0 1

0 0 1

1 0 02

0 1 0

0 0 1

R r

R r r

R r r

−→ − =

−

− = +→ −

= − +−


, ,2 3 4

x y z= − = − = − or

1 2 3, ,

2 3 4− − − .

24. 0

5 6

2 2 1

x y z

x y z

x y z

− + =− − =− + =


1 1 5 6

2 2 1 1

−− −−

2 1 2

3 1 3

1 1 1 0

0 0 6 6 2

0 0 1 1

R r r

R r r

−= − +

→ −= − +

−

( )12 26

1 1 1 0

0 0 1 1

0 0 1 1

R r

−→ − = −

−

1 2 1

3 2 3

1 1 0 1

0 0 1 1

0 0 0 0

R r r

R r r

−= − +

→ −= +

The system is dependent. 1

1

x y

z

= += −

The solution is 1x y= + , 1z = − , y is any real

number or {( , , ) 1, 1, is anyx y z x y z y= + = −

}real number .

25.

1

2 2 3

2 2 3 0

3 4 5 3

x y z t

x y z t

x y z t

x y z t

− − − =+ − + =− − − =− + + = −

Write the augmented matrix: 1 1 1 1 1

2 1 1 2 3

1 2 2 3 0

3 4 1 5 3

− − −−

− − −− −

2 1 2

3 1 3

4 1 4

2 3

2

1 1 1 1 12

0 3 1 4 1

0 1 1 2 13

0 1 4 8 6

1 1 1 1 1

Interchange0 1 1 2 1

and 0 3 1 4 1

0 1 4 8 6

1 1 1 1 1

0 1 1 2 1

0 3 1 4 1

0 1 4 8 6

R r r

R r r

R r r

r r

R

− − −= − +

→ = − +− − − −

= − +− −− − −− − − −

→

− −− − −

→ = −

− −

( )2r

1 2 1

3 2 3

4 2 4

13 32

14 45

2 3 2

4

1 0 0 1 2

0 1 1 2 1 3

0 0 2 2 2

0 0 5 10 5

1 0 0 1 2

0 1 1 2 1

0 0 1 1 1

0 0 1 2 1

1 0 0 1 2=0 1 0 1 0

0 0 1 1 1 =

0 0 0 1 2

R r r

R r r

R r r

R r

R r

R r r

R r

= +→ = − +

− − −= +

−

= −→

=−

− +→

−−

3 4r+

1 4 1

2 4 2

3 4 3

1 0 0 0 4

0 1 0 0 2

0 0 1 0 3

0 0 0 1 2

R r r

R r r

R r r

= − +→ = − +

= − +−

The solution is 4, 2, 3, 2x y z t= = = = − or

(4, 2, 3, 2)− .



26. 3 4

3(3) 4(1) 9 4 51 3

= − = − =

27. 01 4

6 62 1 1 261 2 1 4 03 31 4 4 134 1

1(6 6) 4( 3 24) 0( 1 8)

1(0) 4( 27) 0( 9) 0 108 0

108

− −− = − +

= − − − − + − −= − − + − = + +=

28. 32 1

0 5 5 01 15 0 2 1 ( 3)16 0 0 62 26 02

2(0 6) 1(0 2) 3(30 0)

2( 6) 1( 2) 3(30)

12 2 90

100

−= − + −

= − − − − −= − − − −= − + −= −

29. 2 4

3 2 4x yx y

− =+ =

Set up and evaluate the determinants to use Cramer’s Rule:

1 23 2

1(2) 3( 2) 2 6 8D−= = − − = + =

4 24 2

4(2) 4( 2) 8 8 16xD−= = − − = + =

1 43 4

1(4) 3(4) 4 12 8yD = = − = − = −

The solution is 168

2xDD

x = = = , 88

1yD

Dy −= = = −

or (2, 1)− .

30. 2 3 13 0

3 2 0x y

x y+ − =

− =

Write the system is standard form: 2 3 133 2 0

x yx y

+ =− =


2 34 9 13

3 2D = = − − = −

−

13 326 0 26

0 2xD = = − − = −−

132 0 39 393 0

yD = = − = −

The solution is 2613

2xDD

x −−

= = = ,

3913

3yD

Dy −

−= = = or (2, 3).

31. 2 6

2 3 13

3 2 3 16

x y z

x y z

x y z

+ − =− + = −− + = −


( )

1 2 1

2 1 3

3 2 3

1 3 1 3 2 11 2 ( 1)

2 3 2 3 3 2

1 3 6 2( 3 6) ( 1)( 4 3)

3 6 1 10

D

−= −

−− − −

= − + −− − −

= − + − − + + − − += + + =

( )

6 2 1

13 1 3

16 2 3

1 3 13 3 13 16 2 ( 1)

2 3 16 3 16 2

6 3 6 2( 39 48) ( 1)(26 16)

18 18 10 10

xD

−= − −

− −− − − −

= − + −− − − −

= − + − − + + − −= − − = −

( )

1 6 1

2 13 3

3 16 3

13 3 2 3 2 131 6 ( 1)

16 3 3 3 3 16

1 39 48 6(6 9) ( 1)( 32 39)

9 18 7 20

yD

−= −

−− −

= − + −− −

= − + − − + − − += + − =

( )

1 2 6

2 1 13

3 2 16

1 13 2 13 2 11 2 6

2 16 3 16 3 2

1 16 26 2( 32 39) 6( 4 3)

10 14 6 30

zD = − −− −

− − − −= − +

− − − −= − − − + + − += − − − = −

The solution is 10

110

xDx

D−= = = − ,

202

10yD

yD

= = = , 30

310

zDz

D−= = = − or

( 1, 2, 3)− − .


1171


32. Let 8x y

a b= .

Then ( )2 2 8 162

x y

a b= = by Theorem (14).

The value of the determinant is multiplied by k when the elements of a column are multiplied by k.

33. Let 8x y

a b= .

Then 8y x

b a= − by Theorem (11). The

value of the determinant changes sign when any 2 columns are interchanged.


6

( 4) ( 4)( 4) 4

6 ( 4)

A Bx x x x

x x x x

A x Bx

− = − +− −

= − +

Let x =4, then 6 (4 4) (4)

4 6

32

A B

B

B

= − +=

=

Let x = 0, then 6 (0 4) (0)

4 6

32

A B

A

A

= − +− =

= −

3 36 2 2

( 4) 4x x x x

−= +

− −


2 2

41( 1)

x A B Cx xx x x

− = + +−−

Multiply both sides by 2 ( 1)x x − 24 ( 1) ( 1)x Ax x B x Cx− = − + − +

Let 1x = , then 21 4 (1)(1 1) (1 1) (1)

3

3

A B C

C

C

− = − + − +− =

= −

Let 0x = , then 20 4 (0)(0 1) (0 1) (0)

4

4

A B C

B

B

− = − + − +− = −

=

Let 2x = , then

22 4 (2)(2 1) (2 1) (2)

2 2 4

2 2 4 4( 3)

2 6

3

A B C

A B C

A

A

A

− = − + − +− = + +

= − − − −==

2 2

4 3 4 31( 1)

xx xx x x

− −= + +−−


2 21( 9)( 1) 9

x A Bx Cxx x x

+= +++ + +

Multiply both sides by 2( 1)( 9)x x+ + . 2( 9) ( )( 1)x A x Bx C x= + + + +

Let 1x = − , then

( )( ) ( )( )( )21 1 9 1 1 1

1 (10) ( )(0)

1 10

110

A B C

A B C

A

A

− = − + + − + − +

− = + − +− =

= −

Let 0x = , then

( ) ( )( )( )20 0 9 0 0 1

0 9

10 9

109

10

A B C

A C

C

C

= + + + +

= +

= − +

=

Let 1x = , then

( ) ( )( )( )21 1 9 1 1 1

1 (10) ( )(2)

1 10 2 2

1 91 10 2 2

10 109

1 1 25

A B C

A B C

A B C

B

B

= + + + +

= + += + +

= − + +

= − + +

1

251

10

B

B

=

=

2 2

1 1 910 10 10

1( 9)( 1) 9

xxxx x x

− += +

++ + +




2 2 2 2 2( 4) 4 ( 4)

x Ax B Cx D

x x x

+ += ++ + +

Multiply both sides by 2 2( 4)x + . 3 2

3 3 2

3 3 2

( )( 4)

4 4

(4 ) 4

x Ax B x Cx D

x Ax Bx Ax B Cx D

x Ax Bx A C x B D

= + + + +

= + + + + +

= + + + + +

1; 0A B= = 4 0

4(1) 0

4

A C

C

C

+ =+ =

= −

4 0

4(0) 0

0

B D

D

D

+ =+ =

=

3

2 2 2 2 2

4

( 4) 4 ( 4)

x x x

x x x

−= ++ + +


2 2 2

2

( 1)( 1) ( 1)( 1)( 1)

1 1 1

x x

x x x x x

A B Cx Dx x x

=+ − + − +

+= + +− + +

Multiply both sides by 2( 1)( 1)( 1)x x x− + + . 2 2 2( 1)( 1) ( 1)( 1)

( )( 1)( 1)

x A x x B x x

Cx D x x

= + + + − ++ + − +

Let 1x = , then 2 2 21 (1 1)(1 1) (1 1)(1 1)

( (1) )(1 1)(1 1)

1 4

14

A B

C D

A

A

= + + + − ++ + − +

=

=

Let 1x = − , then 2 2

2

( 1) ( 1 1)(( 1) 1)

( 1 1)(( 1) 1)

( ( 1) )( 1 1)( 1 1)

1 4

14

A

B

C D

B

B

− = − + − +

+ − − − ++ − + − − − +

= −

= −

Let 0x = , then

2 2 20 (0 1)(0 1) (0 1)(0 1)

( (0) )(0 1)(0 1)

0

1 10

4 412

A B

C D

A B D

D

D

= + + + − ++ + − +

= − −

= − − −

=

Let 2x = , then 2 2 22 (2 1)(2 1) (2 1)(2 1)

( (2) )(2 1)(2 1)

4 15 5 6 3

1 1 14 15 5 6 3

4 4 215 5 3

6 44 4 2

6 0

0

A B

C D

A B C D

C

C

C

C

= + + + − ++ + − +

= + + +

= + − + +

= − + −

==

( )( )2

22 2

1 1 14 4 2

1 1 11 1

xx x xx x

−= + +

− + ++ −

39. Solve the first equation for y, substitute into the second equation and solve:

2 2

2 3 0 2 3

5

x y y x

x y

+ + = → = − −+ =

2 2 2 2

2

( 2 3) 5 4 12 9 5

5 12 4 0 (5 2)( 2) 0

2 or 2

511

15

x x x x x

x x x x

x x

y y

+ − − = → + + + =+ + = → + + =

= − = −

= − =

Solutions: 2 11

, , ( 2,1)5 5

− − − .

40. Multiply each side of the second equation by 2 and add the equations to eliminate xy:

2 2

22 2

2

2

2 10 2 10

3 2 2 6 4

7 14

2

2

xy y xy y

xy y xy y

y

y

y

+ = → + =

− + = → − + =

==

= ±


1173


( ) ( )

( ) ( )

2

2

If 2 :

2 2 2 10 2 2 8 2 2

If 2 :

2 2 2 10 2 2 8

2 2

y

x x x

y

x x

x

=

+ = → = → =

= −

− + − = → − =

→ = −

Solutions: ( ) ( )2 2, 2 , 2 2, 2− −

41. Substitute into the second equation into the first equation and solve:

2 2

2

6

3

x y y

x y

+ ==

2

2

3 6

3 0

( 3) 0 0 or 3

y y y

y y

y y y y

+ =− =− = → = =

2 2

2 2

If 0 : 3(0) 0 0

If 3 : 3(3) 9 3

y x x x

y x x x

= = → = → == = → = → = ±

Solutions: (0, 0), (–3, 3), (3, 3).

42. Factor the second equation, solve for x, substitute into the first equation and solve:

2 2

2 2

3 4 5 8

3 2 0

x xy y

x xy y

+ + =+ + =

2 23 2 0

( 2 )( ) 0 2 or

x xy y

x y x y x y x y

+ + =+ + = → = − = −

Substitute 2x y= − and solve: 2 2

2 2

2 2 2

2

2

3 4 5 8

3( 2 ) 4( 2 ) 5 8

12 8 5 8

9 8

8 2 29 3

x xy y

y y y y

y y y

y

y y

+ + =− + − + =

− + ==

= = ±

Substitute x y= − and solve:

2 2

2 2

2 2 2

2

2

3 4 5 8

3( ) 4( ) 5 8

3 4 5 8

4 8

2 2

x xy y

y y y y

y y y

y

y y

+ + =− + − + =

− + ==

= = ±

2 2 2 2 4 2If : 2

3 3 3

2 2 2 2 4 2If : 2

3 3 3

y x

y x

−= = − =

− −= = − =

If 2 : 2

If 2 : 2

y x

y x

= = −

= − =

Solutions:

( )( )

4 2 2 2 4 2 2 2, , , , 2, 2 ,

3 3 3 3

2, 2

− − −

−

43.

2 2

2

3 2

1 0

x x y y

x xy

y

− + + = −− + + =

Multiply each side of the second equation by –y and add the equations to eliminate y:

2 2

2 2

3 2

0

2 2

1

x x y y

x x y y

x

x

− + + = −− + − − =

− = −=

If 1:x = 2 2

2

1 3(1) 2

0

( 1) 0

0 or 1

y y

y y

y y

y y

− + + = −+ =+ =

= = − Note that 0y ≠ because that would cause division by zero in the original system. Solution: (1, –1)

44. 3 4 12x y+ ≤ Graph the line 3 4 12x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since

( ) ( )3 0 4 0 12+ ≤ is true, shade the side of the line

containing (0, 0).

y

x−5

3 + 4 x y 12≤

5

5

−5



45. 2y x≤

Graph the parabola 2y x= . Use a solid curve since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 1). Since 20 1≤ is false, shade the opposite side of the parabola from (0, 1).

y

x−5 5

5

−5

2y x≤

46. 2 2

2

x y

x y

− + ≤+ ≥

Graph the line 2 2x y− + = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since

2(0) 0 2− + ≤ is true, shade the side of the line containing (0, 0). Graph the line 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

y

–5

5

x–5 5

–2x + y = 2

x + y = 2

The graph is unbounded. Find the vertices: To find the intersection of 2x y+ = and

2 2x y− + = , solve the system:

2

2 2

x y

x y

+ =− + =

Solve the first equation for x: 2x y= − . Substitute and solve:

2(2 ) 2

4 2 2

3 6

2

y y

y y

y

y

− − + =− + + =

==

2 2 0x = − =

The point of intersection is (0, 2). The corner point is (0, 2).

47.

0

0

4

2 3 6

x

y

x y

x y

≥≥

+ ≤+ ≤

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 4x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 4+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 3 6x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≤ is true, shade the side of the line containing (0, 0).

y

–2

8

x8

(0, 2)

(3, 0)(0, 0)

x + y = 4

2x + 3y = 6

The overlapping region is the solution. The graph is bounded. Find the vertices: The x-axis and y-axis intersect at (0, 0). The intersection of 2 3 6x y+ = and the y-axis is (0, 2). The intersection of 2 3 6x y+ = and the x-axis is (3, 0). The three corner points are (0, 0), (0, 2), and (3, 0).

48.

0

0

2 8

2 2

x

y

x y

x y

≥≥

+ ≤+ ≥

Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 8x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 8 is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 + 2(0) 2 is false, shade the opposite side of the line from (0, 0).


1175


y

–1

9

x–1 9

2x + y = 8

x + 2y = 2

(2, 0)

(4, 0)(0, 1)

(0, 8)

The overlapping region is the solution. The graph

is bounded. Find the vertices: The intersection of 2 2x y+ = and the y-axis is (0, 1). The

intersection of 2 2x y+ = and the x-axis is (2, 0). The intersection of 2 8x y+ = and the y-axis is (0, 8). The intersection of 2 8x y+ = and the x-axis is (4, 0). The four corner points are (0, 1), (0, 8), (2, 0), and (4, 0).

49. Graph the system of inequalities: 2 2 16

2

x y

x y

+ ≤+ ≥

Graph the circle 2 2 16x y+ = .Use a solid line since the inequality uses . Choose a test point not on the circle, such as (0, 0). Since

2 20 0 16+ ≤ is true, shade the side of the circle containing (0, 0). Graph the line 2x y+ = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

y

–5

5

x–5 5

x + y = 2

x2+ y2 = 16

50. Graph the system of inequalities: 2 2 25

4

x y

xy

+ ≤≤

Graph the circle 2 2 25x y+ = .Use a solid line since the inequality uses . Choose a test point not on the circle, such as (0, 0). Since

2 20 0 25+ ≤ is true, shade the side of the circle containing (0, 0). Graph the hyperbola 4xy = . Use a solid line

since the inequality uses . Choose a test point not on the hyperbola, such as (1, 2). Since 1 2 4⋅ ≤ is true, shade the same side of the hyperbola as (1, 2). The overlapping region is the solution.

51. Maximize 3 4z x y= + subject to 0x ≥ , 0y ≥ , 3 2 6x y+ ≥ , 8x y+ ≤ . Graph the constraints.

y

x

(0,8)

(8,0)(2,0)

(0,3)



Vertex Value of 3 4

(0, 3) 3(0) 4(3) 12

(0, 8) 3(0) 4(8) 32

(2, 0) 3(2) 4(0) 6

(8, 0) 3(8) 4(0) 24

z x y

z

z

z

z

= += + == + == + == + =


52. Minimize 3 5z x y= + subject to 0x ≥ , 0y ≥ , 1x y+ ≥ , 3 2 12x y+ ≤ , 3 12x y+ ≤ .


y

x(4,0)(1,0)

(0,1)

(0,4)127 , 24

7( )



To find the intersection of 3 2 12 and 3 12x y x y+ = + = , solve the system:

3 2 12

3 12

x y

x y

+ =+ =

Solve the second equation for x: 12 3x y= − Substitute and solve: 3(12 3 ) 2 12

36 9 2 12

7 24

247

y y

y y

y

y

− + =− + =

− = −

=

24 72 1212 3 12

7 7 7x = − = − =


,7 7

.


,7 7

.


Vertex Value of 3 5

(0, 1) 3(0) 5(1) 5

(0, 4) 3(0) 5(4) 20

(1, 0) 3(1) 5(0) 3

(4, 0) 3(4) 5(0) 12

12 24 12 24 156, 3 5

7 7 7 7 7

z x y

z

z

z

z

z

= += + =

= + == + =

= + =

= + =


53. 2 5 5

4 10

x y

x y A

+ =+ =

Multiply each side of the first equation by –2 and eliminate x:

4 10 10

4 10

0 10

x y

x y A

A

− − = −+ =

= −

If there are to be infinitely many solutions, the result of elimination should be 0 = 0. Therefore,

10 0 or 10A A− = = .

54. 2 5 5

4 10

x y

x y A

+ =+ =

Multiply each side of the first equation by –2 and eliminate x:

4 10 10

4 10

0 10

x y

x y A

A

− − = −+ =

= −

If the system is to be inconsistent, the result of elimination should be 0 = any number except 0. Therefore, 10 0 or 10A A− ≠ ≠ .

55. 2y ax bx c= + + At (0, 1) the equation becomes:

21 (0) (0)

1

a b c

c

= + +=


0

0

a b c

a b c

a b c

= + += + +

+ + =

At (–2, 1) the equation becomes: 21 ( 2) ( 2)

1 4 2

4 2 1

a b c

a b c

a b c

= − + − += − +

− + =


4 2 1

1

a b c

a b c

c

+ + =− + =

=

Substitute 1c = into the first and second equations and simplify:

1 0

1

1

a b

a b

a b

+ + =+ = −

= − −

4 2 1 1

4 2 0

a b

a b

− + =− =

Solve the first equation for a, substitute into the second equation and solve:

4( 1) 2 0

4 4 2 0

6 4

23

b b

b b

b

b

− − − =− − − =

− =

= −

2 11

3 3a = − = −

The quadratic function is 21 21

3 3y x x= − − + .

56. Let x = the number of pounds of coffee that costs $6.00 per pound, and let y = the number of pounds of coffee that costs $9.00 per pound. Then 100x y+ = represents the total amount of


1177


coffee in the blend. The value of the blend will be represented by the equation: 6 9 6.90(100)x y+ = . Solve the system of equations:

100

6 9 690

x y

x y

+ =+ =

Solve the first equation for y: 100y x= − . Solve by substitution: 6 9(100 ) 690

6 900 9 690

3 210

70

x x

x x

x

x

+ − =+ − =

− = −=

100 70 30y = − = The blend is made up of 70 pounds of the $6.00-per-pound coffee and 30 pounds of the $9.00-per-pound coffee.

57. Let x = the number of small boxes, let y = the number of medium boxes, and let z = the number of large boxes. Oatmeal raisin equation: 2 2 15x y z+ + = Chocolate chip equation: 2 10x y z+ + = Shortbread equation: 3 11y z+ =

2 2 15

2 10

3 11

x y z

x y z

y z

+ + =+ + =

+ =

Multiply each side of the second equation by –1 and add to the first equation to eliminate x:

2 2 15

2 10

3 11

5

x y z

x y z

y z

y

+ + =− − − = −

+ ==

Substituting and solving for the other variables: 5 3 11

3 6

2

z

z

z

+ ===

5 2(2) 10

9 10

1

x

x

x

+ + =+ =

=

Thus, 1 small box, 5 medium boxes, and 2 large boxes of cookies should be purchased.

58. a. Let x = the number of lower-priced packages, and let y = the number of quality packages. Peanut inequality: 8 6 120(16)

4 3 960

x y

x y

+ ≤+ ≤

Cashew inequality: 4 6 72(16)

2 3 576

x y

x y

+ ≤+ ≤

The system of inequalities is:

0

0

4 3 960

2 3 576

x

y

x y

x y

≥≥

+ ≤+ ≤

b. Graphing:

To find the intersection of 2 3 576x y+ =

and 4 3 960x y+ = , solve the system:

4 3 960

2 3 576

x y

x y

+ =+ =

Subtract the second equation from the first: 4 3 960

2 3 576

2 384

192

x y

x y

x

x

+ =− − =

==

Substitute and solve: 2(192) 3 576

3 192

64

y

y

y

+ ===

The corner points are (0, 0), (0, 192), (240, 0), and (192, 64).

59. Let x = the speed of the boat in still water, and

let y = the speed of the river current. The distance from Chiritza to the Flotel Orellana is 100 kilometers.

Rate Time Distance

trip downstream 5 / 2 100

trip downstream 3 100

x y

x y

+−

The system of equations is:



5( ) 100

23( ) 100

x y

x y

+ =

− =

Multiply both sides of the first equation by 6, multiply both sides of the second equation by 5, and add the results.

15 15 600

15 15 500

30 1100

x y

x y

x

+ =− =

=

1100 110

30 3x = =

1103 3 100

3110 3 100

10 3

103

y

y

y

y

− =

− ==

=

The speed of the boat is 110 / 3 36.67 km/hr≈ ; the speed of the current is 10 / 3 3.33 km/hr≈ .

60. Let x = the number of hours for Bruce to do the job alone, let y = the number of hours for Bryce to do the job alone, and let z = the number of hours for Marty to do the job alone. Then 1/x represents the fraction of the job that Bruce does in one hour.

1/y represents the fraction of the job that Bryce does in one hour. 1/z represents the fraction of the job that Marty does in one hour. The equation representing Bruce and Bryce working together is:

( )1 1 1 3

0.754 / 3 4x y

+ = = =

The equation representing Bryce and Marty working together is:

( )1 1 1 5

0.6258 / 5 6y z

+ = = =

The equation representing Bruce and Marty working together is:

( )1 1 1 3

0.3758 / 3 8x z

+ = = =

Solve the system of equations: 1 1

1 1

1 1

0.75

0.625

0.375

x y

y z

x z

− −

− −

− −

+ =

+ =

+ =

Let 1 1 1, ,u x v y w z− − −= = =

0.75

0.625

0.375

u v

v w

u w

+ =+ =+ =

Solve the first equation for u: 0.75u v= − . Solve the second equation for w: 0.625w v= − . Substitute into the third equation and solve: (0.75 ) (0.625 ) 0.375

2 1

0.5

v v

v

v

− + − =− = −

=

0.75 0.5 0.25u = − = 0.625 0.5 0.125w = − =

Solve for , , andx y z : 4, 2, 8 (reciprocals)x y z= = =

Bruce can do the job in 4 hours, Bryce in 2 hours, and Marty in 8 hours.

61. Let x = the number of gasoline engines produced each week, and let y = the number of diesel engines produced each week. The total cost is:

450 550C x y= + . Cost is to be minimized; thus, this is the objective function. The constraints are:

20 60x≤ ≤ number of gasoline engines needed and capacity each week.

15 40y≤ ≤ number of diesel engines needed and capacity each week.

50x y+ ≥ number of engines produced to prevent layoffs.

Chapter 11 Test

1179



y

x

(60,40)

(60,15)

(20,40)

(35,15)

(20,30)


(60, 15), (60, 40) Evaluate the objective function:

( ) ( ) ( )


(20, 30) 450(20) 550(30) 25,500

(20, 40) 450(35) 550(40) 31,000

(35, 15) 450(35) 550(15) 24,000

(60, 15) 450(60) 550(15) 35,250

60, 40 450 60 550 40 49,000

C x y

C

C

C

C

C

= += + == + == + == + =

= + =

The minimum cost is $24,000, when 35 gasoline engines and 15 diesel engines are produced. The excess capacity is 15 gasoline engines, since only 20 gasoline engines had to be delivered.


Chapter 11 Test

1. 2 7

4 3 9

x y

x y

− + = −+ =

Substitution: We solve the first equation for y, obtaining

2 7y x= − Next we substitute this result for y in the second equation and solve for x.

( )4 3 9

4 3 2 7 9

4 6 21 9

10 30

303

10

x y

x x

x x

x

x

+ =+ − =

+ − ==

= =

We can now obtain the value for y by letting 3x = in our substitution for y.

( )2 7

2 3 7 6 7 1

y x

y

= −= − = − = −

The solution of the system is 3x = , 1y = − or

(3, 1)− .

Elimination: Multiply each side of the first equation by 2 so that the coefficients of x in the two equations are negatives of each other. The result is the equivalent system

4 2 14

4 3 9

x y

x y

− + = −+ =

We can replace the second equation of this system by the sum of the two equations. The result is the equivalent system

4 2 14

5 5

x y

y

− + = −= −

Now we solve the second equation for y. 5 5

51

5

y

y

= −−= = −

We back-substitute this value for y into the original first equation and solve for x.

( )2 7

2 1 7

2 6

63

2

x y

x

x

x

− + = −− + − = −

− = −−= =−

The solution of the system is 3x = , 1y = − or

(3, 1)− .

2. 1

2 13

5 30 18

x y

x y

− =

− =

We choose to use the method of elimination and multiply the first equation by 15− to obtain the equivalent system

5 30 15

5 30 18

x y

x y

− + = −− =

We replace the second equation by the sum of the two equations to obtain the equivalent system

5 30 15

0 3

x y− + = −=

The second equation is a contradiction and has no solution. This means that the system itself has no solution and is therefore inconsistent.


1180


3. 2 5 (1)

3 4 2 (2)

5 2 3 8 (3)

x y z

x y z

x y z

− + =+ − = −

+ + =

We use the method of elimination and begin by eliminating the variable y from equation (2). Multiply each side of equation (1) by 4 and add the result to equation (2). This result becomes our new equation (2). 2 5 4 4 8 20

3 4 2 3 4 2

7 7 18 (2)

x y z x y z

x y z x y z

x z

− + = − + =+ − = − + − = −

+ =

We now eliminate the variable y from equation (3) by multiplying each side of equation (1) by 2 and adding the result to equation (3). The result becomes our new equation (3). 2 5 2 2 4 10

5 2 3 8 5 2 3 8

7 7 18 (3)

x y z x y z

x y z x y z

x z

− + = − + =+ + = + + =

+ =

Our (equivalent) system now looks like 2 5 (1)

7 7 18 (2)

7 7 18 (3)

x y z

x z

x z

− + =+ =+ =

Treat equations (2) and (3) as a system of two equations containing two variables, and eliminate the x variable by multiplying each side of equation (2) by 1− and adding the result to equation (3). The result becomes our new equation (3). 7 7 18 7 7 18

7 7 18 7 7 18

0 0 (3)

x z x z

x z x z

+ = − − = −+ = + =

=

We now have the equivalent system 2 5 (1)

7 7 18 (2)

0 0 (3)

x y z

x z

− + =+ =

=

This is equivalent to a system of two equations with three variables. Since one of the equations contains three variables and one contains only two variables, the system will be dependent. There are infinitely many solutions. We solve equation (2) for x and determine that

187

x z= − + . Substitute this expression into

equation (1) to obtain y in terms of z.

2 5

182 5

718

2 57

177

177

x y z

z y z

z y z

y z

y z

− + =

− + − + =

− + − + =

− + =

= −


,7 7

x z y z= − + = − ,

z is any real number or 18

( , , ) ,7

x y z x z= − +

17, is any real number

7y z z= − .

4. 23

3 2 8 3 (1)

1 (2)

6 3 15 8 (3)

x y z

x y z

x y z

+ − = −− − + =

− + =

We start by clearing the fraction in equation (2) by multiplying both sides of the equation by 3.

3 2 8 3 (1)

3 2 3 3 (2)

6 3 15 8 (3)

x y z

x y z

x y z

+ − = −− − + =

− + =

We use the method of elimination and begin by eliminating the variable x from equation (2). The coefficients on x in equations (1) and (2) are negatives of each other so we simply add the two equations together. This result becomes our new equation (2). 3 2 8 3

3 2 3 3

5 0 (2)

x y z

x y z

z

+ − = −− − + =

− =

We now eliminate the variable x from equation (3) by multiplying each side of equation (1) by

2− and adding the result to equation (3). The result becomes our new equation (3). 3 2 8 3 6 4 16 6

6 3 15 8 6 3 15 8

7 31 14 (3)

x y z x y z

x y z x y z

y z

+ − = − − − + =− + = − + =

− + =

Our (equivalent) system now looks like 3 2 8 3 (1)

5 0 (2)7 31 14 (3)

x y zz

y z

+ − = −− =

− + =

We solve equation (2) for z by dividing both

Chapter 11 Test

1181


sides of the equation by 5− . 5 0

0

z

z

− ==

Back-substitute 0z = into equation (3) and solve for y.

7 31 147 31(0) 14

7 142

y zy

yy

− + =− + =

− == −

Finally, back-substitute 2y = − and 0z = into equation (1) and solve for x.

3 2 8 33 2( 2) 8(0) 3

3 4 33 1

13

x y zx

xx

x

+ − = −+ − − = −

− = −=

=

The solution of the original system is 13

x = , 2y = − , 0z = or 1

, 2, 03

− .

5. 4 5 0

2 6 19

5 5 10

x y z

x y

x y z

− + =− − + = −

+ − =

We first check the equations to make sure that all variable terms are on the left side of the equation and the constants are on the right side. If a variable is missing, we put it in with a coefficient of 0. Our system can be rewritten as

4 5 0

2 0 25

5 5 10

x y z

x y z

x y z

− + =− − + = −

+ − =

The augmented matrix is 4 5 1 02 1 0 25

1 5 5 10

−− − −

−

6. The matrix has three rows and represents a system with three equations. The three columns to the left of the vertical bar indicate that the system has three variables. We can let x, y, and z denote these variables. The column to the right of the vertical bar represents the constants on the right side of the equations. The system is

3 2 4 6

1 0 8 2

2 1 3 11

x y z

x y z

x y z

+ + = −+ + =

− + + = − or

3 2 4 6

8 2

2 3 11

x y z

x z

x y z

+ + = −+ =

− + + = −

7. 1 1 4 6

2 2 0 4 1 3

3 2 1 8

2 2 4 6 6 4

0 8 1 3 1 11

6 4 1 8 5 12

A C

−+ = − + −

−−

= − + − = −−

8. 1 1 4 6

3 0 4 3 1 3

3 2 1 8

1 1 12 18 11 19

0 4 3 9 3 5

3 2 3 24 6 22

A C

−− = − − −

−− − −

= − − − = −− −

9. Here we are taking the product of a 3 2× matrix and a 2 3× matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (2 in both cases), the operation can be performed and will result in a 3 3× matrix.

4 1 6 0 4( 2) 6 3 4 5 6 1

1 1 ( 3)0 1( 2) ( 3)3 1 5 ( 3)1

( 1)1 8 0 ( 1)( 2) 8 3 ( 1)5 8 1

4 10 26

1 11 2

1 26 3

4 61 2 5

1 30 3 1

1 8

CB

⋅ + ⋅ − + ⋅ ⋅ + ⋅

⋅ + − − + − ⋅ + −

− + ⋅ − − + ⋅ − + ⋅

−

−

−= −

−

=

=

10. Here we are taking the product of a 2 3× matrix and a 3 2× matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (3 in both cases), the operation can be performed and will result in a 2 2× matrix.

( ) ( ) ( ) ( )( ) ( )

1 1 2 0 5 3 1 1 2 4 5 2

0 1 3 0 1 3 0 1 3 4 1 2

1 11 2 5

0 40 3 1

3 2

16 17

3 10

BA

⋅ + − ⋅ + ⋅ ⋅ − + − ⋅ − + ⋅

⋅ + ⋅ + ⋅ ⋅ − + − + ⋅

−−

= −

=

=−


1182


11. We first form the matrix

[ ]2

3 2 1 0|

5 4 0 1A I =

Next we use row operations to transform [ ]2|A I

into reduced row echelon form.

( )1 1

2 113 33

1 03 2 1 0

5 4 0 1 5 4 0 1R r→ =

( )

( )

( )

2 1 2

2 2

1 2 1

2 13 3

523 3

2 13 3 3

25 32 2

25 3 32 2

1 0 5

0 1

1 0

0 1

1 0 2 1

0 1

R r r

R r

R rr

→ = − +−

→ =−

−→ = − +−

Therefore, 15 32 2

2 1A− −

=−

.

12. We first form the matrix

[ ]3

1 1 1 1 0 0

| 2 5 1 0 1 0

2 3 0 0 0 1

B I

−= −

Next we use row operations to transform [ ]3|B I

into reduced row echelon form.

( )

2 1 2

3 1 3

2 2

1 2 1

3

3 2 1 17 7 7 7

54 17 7 73 2 17 7 7

51 47 7 7

1 1 1 1 0 0

2 5 1 0 1 0

2 3 0 0 0 1

1 1 1 1 0 02

0 7 3 2 1 0 2

0 5 2 2 0 1

1 1 1 1 0 0

0 1 0

0 5 2 2 0 1

1 0 0

0 1 0

0 0 1

R r r

R r r

R r

R r r

R

−−

−= − +

→ − −= − +

− −−

→ − − =− −

= +→ − −

− − 2 35r r= − +

( )3 3

1 3 1

32 3 27

54 17 7 73 2 17 7 7

47

1 0 0

0 1 0 7

0 0 1 4 5 7

1 0 0 3 3 4

0 1 0 2 2 3

0 0 1 4 5 7

R r

R r r

R r r

→ − − =− −

− = − +→ − −

= +− −

Thus, 1

3 3 4

2 2 3

4 5 7

B−

−= − −

− −

13. 6 3 12

2 2

x y

x y

+ =− = −

We start by writing the augmented matrix for the system.

6 3 12

2 1 2− −

Next we use row operations to transform the augmented matrix into row echelon form.

( )

( )

1 2

2 1

11 12

2 1 2

12

12

6 3 12 2 1 2

2 1 2 6 3 12

1 1

6 3 12

1 1 6

0 6 18

R r

R r

R r

R r r

=− −→

− − =

− −→ =

− −→ = − +

( )

( )

12 26

12 2 12

12

12

1 1

0 1 3

1 0

0 1 3

R r

R r r

− −→ =

→ = +

Chapter 11 Test

1183


The solution of the system is 12x = , 3y = or

( )12 , 3

14. 1

74

8 2 56

x y

x y

+ =

+ =


141 7

8 2 56


14

2 1 2

14

1 788 2 56

1 7

0 0 0

R R r= − +

The augmented matrix is now in row echelon form. Because the bottom row consists entirely of 0’s, the system actually consists of one equation in two variables. The system is dependent and therefore has an infinite number of solutions. Any ordered pair satisfying the

equation 1

74

x y+ = , or 4 28y x= − + , is a

solution to the system.

15. 2 4 3

2 7 15 12

4 7 13 10

x y z

x y z

x y z

+ + = −+ + = −+ + = −


1 2 4 3

2 7 15 12

4 7 13 10

−−−


( )

2 1 2

3 1 3

2 3

3 2

3 2 3

1 2 4 3

2 7 15 12

4 7 13 10

1 2 4 32

0 3 7 6 4

0 1 3 2

1 2 4 3

0 1 3 2

0 3 7 6

1 2 4 3

0 1 3 2 3

0 0 2 0

1 2 4 3

0 1 3 2

0 0 1 0

R r r

R r r

R r

R r

R r r

−−−

−= − +

→ −= − +

− −−

= −→ −

=−

−→ − = − +

−−

→ − ( )13 32 R r= −

The matrix is now in row echelon form. The last row represents the equation 0z = . Using 0z = we back-substitute into the equation 3 2y z+ = − (from the second row) and obtain

( )3 2

3 0 2

2

y z

y

y

+ = −+ = −

= −

Using 2y = − and 0z = , we back-substitute into the equation 2 4 3x y z+ + = − (from the first row) and obtain

( ) ( )2 4 3

2 2 4 0 3

1

x y z

x

x

+ + = −+ − + = −

=

The solution is 1x = , 2y = − , 0z = or

(1, 2, 0)− .

16. 2 2 3 5

2 8

3 5 8 2

x y z

x y z

x y z

+ − =− + =

+ − = −


2 2 3 5

1 1 2 8

3 5 8 2

−−

− −



1184


( )

1 2

2 1

2 1 2

3 1 3

12 24

7 114 4

7 114 4

2 2 3 5

1 1 2 8

3 5 8 2

1 1 2 8

2 2 3 5

3 5 8 2

1 1 2 82

0 4 7 11 3

0 8 14 26

1 1 2 8

0 1

0 8 14 26

1 1 2 8

0 1

R r

R r

R r r

R r r

R r

−−

− −−

== −

=− −

−= − +

= − −= − +

− −

−= =− −

− −

−= − − ( )3 2 3 8

0 0 0 4

R r r= − +

−

The last row represents the equation 0 4= − which is a contradiction. Therefore, the system has no solution and is be inconsistent.

17. ( ) ( ) ( )( )2 52 7 5 3 14 15 29

3 7

−= − − = − − = −

18. 2 4 6

1 4 0

1 2 4

−

− −

[ ] [ ][ ]

4 0 1 0 1 42 ( 4) 6

2 4 1 4 1 2

2 4( 4) 2(0) 4 1( 4) ( 1)(0)

6 1(2) ( 1)4

2( 16) 4( 4) 6(6)

32 16 36

12

= − − +− − − −

= − − + − − −+ − −

= − + − += − − += −

19. 4 3 23

3 5 19

x y

x y

+ = −− =

The determinant D of the coefficients of the variables is

( )( ) ( )( )4 34 5 3 3 20 9 29

3 5D = = − − = − − = −

−

Since 0D ≠ , Cramer’s Rule can be applied.

( )( ) ( )( )23 323 5 3 19 58

19 5xD−

= = − − − =−

( )( ) ( )( )4 234 19 23 3 145

3 19yD−

= = − − =

582

29xD

xD

= = = −−

1455

29yD

yD

= = = −−

The solution of the system is 2x = − , 5y = − or

( 2, 5)− − .

20. 4 3 2 15

2 3 15

5 5 2 18

x y z

x y z

x y z

− + =− + − = −

− + =

The determinant D of the coefficients of the variables is

( )

( ) ( ) ( )( ) ( ) ( )

4 3 2

2 1 3

5 5 2

1 3 2 3 2 14 3 2

5 2 5 2 5 5

4 2 15 3 4 15 2 10 5

4 13 3 11 2 5

52 33 10

9

D

−= − −

−− − − −

= − − +− −

= − + − + + −= − + += − + += −

Since 0D ≠ , Cramer’s Rule can be applied.

( )

( ) ( ) ( )( ) ( ) ( )

15 3 2

15 1 3

18 5 2

1 3 15 3 15 115 3 2

5 2 18 2 18 5

15 2 15 3 30 54 2 75 18

15 13 3 24 2 57

9

xD

−= − −

−− − − −

= − − +− −

= − + − + + −= − + += −

Chapter 11 Test

1185


( ) ( ) ( )( ) ( ) ( )

4 15 2

2 15 3

5 18 2

15 3 2 3 2 154 15 2

18 2 5 2 5 18

4 30 54 15 4 15 2 36 75

4 24 15 11 2 39

9

yD = − − −

− − − − − −= − +

= − + − − + + − += − += −

( )

( ) ( ) ( )( ) ( ) ( )

4 3 15

2 1 15

5 5 18

1 15 2 15 2 14 3 15

5 18 5 18 5 5

4 18 75 3 36 75 15 10 5

4 57 3 39 15 5

36

zD

−= − −

−− − − −

= − − +− −

= − + − + + −= − + += −

9

19

xDx

D−= = =−

, 9

19

yDy

D= = = −

−,

364

9zD

zD

−= = =−

The solution of the system is 1x = , 1y = − ,

4z = or (1, 1, 4)− .

21. 2 2

2

3 12

9

x y

y x

+ ==

Substitute 9x for 2y into the first equation and solve for x:

( )2

2

2

3 9 12

3 9 12 0

3 4 0

( 1)( 4) 0

x x

x x

x x

x x

+ =

+ − =+ − =

− + =

1 or 4x x= = − Back substitute these values into the second equation to determine y:

1x = : 2 9(1) 9

3

y

y

= == ±

4x = − : 2 9( 4) 36

36 (not real)

y

y

= − = −

= ± −

The solutions of the system are (1, 3)− and

(1, 3) .

22. 2 22 3 5

1 1

y x

y x y x

− =− = = +

Substitute 1x + for y into the first equation and solve for x:

( )( )

2 2

2 2

2 2

2

2

2 1 3 5

2 2 1 3 5

2 4 2 3 5

4 3 0

4 3 0

( 1)( 3) 0

x x

x x x

x x x

x x

x x

x x

+ − =

+ + − =

+ + − =− + − =

− + =− − =

1 or 3x x= = Back substitute these values into the second equation to determine y:

1x = : 1 1 2y = + = 3x = : 3 1 4y = + =

The solutions of the system are (1, 2) and

(3, 4) .

23. 2 2 100

4 3 0

x y

x y

+ ≤− ≥

Graph the circle 2 2 100x y+ = . Use a solid curve since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since

2 20 0 100+ ≤ is true, shade the same side of the circle as (0, 0); that is, inside the circle.

Graph the line 4 3 0x y− = . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 1). Since 4(0) 3(1) 0− ≥ is false, shade the opposite side of the line from (0, 1). The overlapping region is the solution.


1186


24. ( )2

3 7

3

x

x

++

The denominator contains the repeated linear factor 3x + . Thus, the partial fraction decomposition takes on the form

( ) ( )2 2

3 733 3

x A Bxx x

+ = +++ +

Clear the fractions by multiplying both sides by

( )23x + . The result is the identity

( )3 7 3x A x B+ = + +

or ( )3 7 3x Ax A B+ = + +

We equate coefficients of like powers of x to obtain the system

3

7 3

A

A B

== +

Therefore, we have 3A = . Substituting this result into the second equation gives

( )7 3

7 3 3

2

A B

B

B

= += +

− =

Thus, the partial fraction decomposition is

( ) ( )2 2

3 7 3 233 3

xxx x

+ −= +++ +

.

25. ( )

2

22

4 3

3

x

x x

−

+

The denominator contains the linear factor x and the repeated irreducible quadratic factor 2 3x + . The partial fraction decomposition takes on the form

( ) ( )2

2 2 22 2

4 333 3

x A Bx C Dx Ex xx x x

− + += + +++ +

We clear the fractions by multiplying both sides

by ( )22 3x x + to obtain the identity

( ) ( )( ) ( )2 2224 3 3 3x A x x x Bx C x Dx E− = + + + + + +

Collecting like terms yields

( ) ( )( ) ( )

2 4 3 24 3 6 3

3 9

x A B x Cx A B D x

C E x A

− = + + + + ++ + +

Equating coefficients, we obtain the system

0

0

6 3 4

3 0

9 3

A B

C

A B D

C E

A

+ ==

+ + =+ =

= −

From the last equation we get 13

A = − .

Substituting this value into the first equation

gives 13

B = . From the second equation, we

know 0C = . Substituting this value into the fourth equation yields 0E = .

Substituting 13

A = − and 13

B = into the third

equation gives us

( ) ( )1 13 36 3 4

2 1 4

5

D

D

D

− + + =− + + =

=

Therefore, the partial fraction decomposition is

( ) ( ) ( )2

2 222 2

1 13 34 3 5

33 3

xx xx xx x x

−− = + +++ +

26. 0

0

2 8

2 3 2

x

y

x y

x y

≥≥+ ≥

− ≥

The inequalities 0x ≥ and 0y ≥ require that the graph be in quadrant I.

2 8

14

2

x y

y x

+ ≥

≥ − +

Test the point ( )0,0 .

( )2 8

0 2 0 8 ?

0 8 false

x y+ ≥+ ≥≥

The point ( )0,0 is not a solution. Thus, the

graph of the inequality 2 8x y+ ≥ includes the

half-plane above the line 1

42

y x= − + . Because

the inequality is non-strict, the line is also part of the graph of the solution.

Chapter 11 Test

1187


2 3 2

2 23 3

x y

y x

− ≥

≤ −

Test the point ( )0,0 .

( ) ( )2 3 2

2 0 3 0 2 ?

0 2 false

x y− ≥− ≥

≥

The point ( )0,0 is not a solution. Thus, the

graph of the inequality 2 3 2x y− ≥ includes the

half-plane below the line 2 23 3

y x= − .

Because the inequality is non-strict, the line is also part of the graph of the solution. The overlapping shaded region (that is, the shaded region in the graph below) is the solution to the system of linear inequalities.

The graph is unbounded. The corner points are ( )4, 2 and ( )8,0 .

27. The objective function is 5 8z x y= + . We seek the largest value of z that can occur if x and y are solutions of the system of linear inequalities

0

2 8

3 3

x

x y

x y

≥+ ≤

− ≤ −

2 8

2 8

x y

y x

+ == − +

3 3

3 3

11

3

x y

y x

y x

− = −− = − −

= +

The graph of this system (the feasible points) is

shown as the shaded region in the figure below. The corner points of the feasible region are ( )0,1 , ( )3, 2 , and ( )0,8 .

4

8

y

x4 8

2 8x y+ =

3 3x y− = −

(0, 1)(3, 2)

(0, 8)

( )

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

Corner point, , Value of obj. function,

0,1 5 0 8 1 8

3,2 5 3 8 2 31

0,8 5 0 8 8 64

x y z

z

z

z

= + == + == + =

From the table, we can see that the maximum value of z is 64, and it occurs at the point ( )0,8 .

28. Let j = unit price for flare jeans, c = unit price for camisoles, and t = unit price for t-shirts. The given information yields a system of equations with each of the three women yielding an equation.

2 2 4 90 (Megan)

3 42.5 (Paige)

3 2 62 (Kara)

j c t

j t

j c t

+ + =+ =

+ + =

We can solve this system by using matrices.

( )11 12

2 2 4 90 1 1 2 451 0 3 42.5 1 0 3 42.5 1 3 2 62 1 3 2 62

R r= =


1188


( )

( )

2 1 2

3 1 3

2 2

1 2 1

3 2 3

3 312

1 1 2 450 1 1 2.5 0 2 0 17

1 1 2 450 1 1 2.5 0 2 0 17

1 0 3 42.50 1 1 2.5

20 0 2 12

1 0 3 42.50 1 1 2.5 0 0 1 6

R r r

R r r

R r

R r r

R r r

R r

= − += − −

= − +

= − = −

= − += −

= − +

= − =

The last row represents the equation 6z = . Substituting this result into 2.5y z− = (from the second row) gives 2.5

6 2.5

8.5

y z

y

y

− =− =

=

Substituting 6z = into 3 42.5x z+ = (from the first row) gives

( )3 42.5

3 6 42.5

24.5

x z

x

x

+ =+ =

=

Thus, flare jeans cost $24.50, camisoles cost $8.50, and t-shirts cost $6.00.

Chapter 11 Cumulative Review

1. 22 0x x− = ( )2 1 0

0 or 2 1 0

2 1

1

2

x x

x x

x

x

− == − =

=

=

The solution set is 1

0,2

.

2. 3 1 4x + =

( )223 1 4

3 1 16

3 15

5

x

x

x

x

+ =

+ ===

Check:

3 5 1 4

15 1 4

16 4

4 4

⋅ + =

+ =

==

The solution set is { }5 .

3. 3 22 3 8 3 0x x x− − − =

The graph of 3 21 2 3 8 3Y x x x= − − − appears to

have an x-intercept at 3x = .

Using synthetic division:

3 2 3 8 3

6 9 3

2 3 1 0

− − −

Therefore,

( )( )( )( )( )

3 2

2

2 3 8 3 0

3 2 3 1 0

3 2 1 1 0

13 or or 1

2

x x x

x x x

x x x

x x x

− − − =

− + + =

− + + =

= = − = −


1, ,32

− − .

4. 13 9x x+=

( ) 12

2 2

3 3

3 3

2 2

2

xx

x x

x x

x

+

+

=

== += −

The solution set is { }2− .

5. ( ) ( )3 3log 1 log 2 1 2x x− + + =

( )( )( )( )( )

( ) ( )

3

2

2

2

log 1 2 1 2

1 2 1 3

2 1 9

2 10 0

2 5 2 0

5 or 2

2

x x

x x

x x

x x

x x

x x

− + =

− + =

− − =− − =

− + =

= = −

Since 2x = − makes the original logarithms

undefined, the solution set is 52

.

Chapter 11 Cumulative Review

1189


6. 3x e=

( )ln 3 ln

ln 3 1

10.910

ln 3

x e

x

x

=

=

= ≈


0.910ln 3

≈ .

7. 3

4

2( )

1x

g xx

=+

( )( )

( )3 3

4 4

2 2( )

11

x xg x g x

xx

− −− = = = −+− +

Thus, g is an odd function and its graph is symmetric with respect to the origin.

8. 2 2 2 4 11 0x y x y+ − + − = 2 2

2 2

2 2

2 4 11

( 2 1) ( 4 4) 11 1 4

( 1) ( 2) 16

x x y y

x x y y

x y

− + + =− + + + + = + +

− + + =

Center: (1,–2); Radius: 4

9. 2( ) 3 1xf x −= +

Using the graph of 3xy = , shift the graph horizontally 2 units to the right, then shift the graph vertically upward 1 unit.

Domain: ( , )−∞ ∞ Range: (1, )∞

Horizontal Asymptote: 1y =

10. 5

( )2

f xx

=+

52

5 Inverse

2( 2) 5

2 55 25 2 5

2

yx

xy

x yxy x

xy xx

yx x

=+

=+

+ =+ =

= −−= = −

Thus, 1 5( ) 2f x

x− = −

Domain of f = { | 2}x x ≠ −

Range of f = { | 0}y y ≠

Domain of 1f − = { | 0}x x ≠

Range of 1f − = { | 2}y y ≠ − .

11. a. 3 6y x= + The graph is a line. x-intercept: y-intercept:

0 3 6

3 6

2

x

x

x

= += −= −

( )3 0 6

6

y = +=

b. 2 2 4x y+ = The graph is a circle with center (0, 0) and


1190


radius 2.

c. 3y x=

d. 1

yx

=

e. y x=

f. xy e=

g. lny x=

h. 2 22 5 1x y+ = The graph is an ellipse.

2 2

2 2

1 12 5

2 52 5

1

1

x y

x y

+ =

+ =

Chapter 11 Projects

1191


i. 2 23 1x y− = The graph is a hyperbola

2 2

2 2

13

11 3

3

11

x y

x y

− =

− =

j. 2

2

2

2

2 4 1 0

2 1 4

4 ( 1)

1( 1)

4

x x y

x x y

y x

y x

− − + =− + =

= −

= −

12. 3( ) 3 5f x x x= − +

a. Let 31 3 5Y x x= − + .

−3

−6

9

6

The zero of f is approximately 2.28− .

b.

−3

−6

9

6

−3

−6

9

6

f has a local maximum of 7 at 1x = − and a

local minimum of 3 at 1x = .

c. f is increasing on the intervals ( , 1)−∞ − and

(1, )∞ .

Chapter 11 Projects

Project I – Internet Based Project

1. 80% = 0.80 18% = 0.18 2% = 0.02 40% = 0.40 50% = 0.50 10% = 0.10 20% = 0.20 60% = 0.60 20% = 0.20

2. 0.80 0.18 0.02

0.40 0.50 0.10

0.20 0.60 0.20

3. 0.80 0.18 0.02 1.00

0.40 0.50 0.10 1.00

0.20 0.60 0.20 1.00

+ + =+ + =+ + =

The sum of each row is 1 (or 100%). These represent the three possibilities of educational achievement for a parent of a child, unless someone does not attend school at all. Since these are rounded percents, chances are the other possibilities are negligible.

4.

2

2

0.8 0.18 0.02

0.4 0.5 0.1

0.2 0.6 0.2

0.716 0.246 0.038

0.54 0.382 0.078

0.44 0.456 0.104

P =

=

Grandchild of a college graduate is a college graduate: entry (1, 1): 0.716. The probability is 71.6%


1192


5. Grandchild of a high school graduate finishes college: entry (2,1): 0.54. The probability is 54%.

6. grandchildren k = 2. (2) (0) 2

0.716 0.246 0.038

[0.288 0.569 0.143] 0.54 0.382 0.078

0.44 0.456 0.104

[0.576388 0.353414 0.070198]

v v P=

=

=College: 57.6%≈ High School: 35%≈ Elementary: 7%≈

7. The matrix totally stops changing at

30

0.64885496 0.29770992 0.05343511

0.64885496 0.29770992 0.05343511

0.64885496 0.29770992 0.05343511

P ≈

Project II

a. 2 2 2 2 16× × × = codewords.

b. v uG= u will be the matrix representing all of the 4-digit information bit sequences.

0 0 0 0

0 0 0 1

0 0 1 0

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1

u =

(Remember, this is mod two. That means that you only write down the remainder when dividing by 2. )

0 0 0 0 0 0 0

0 0 0 1 1 1 0

0 0 1 0 1 0 1

0 0 1 1 0 1 1

0 1 0 0 0 1 1

0 1 0 1 1 0 1

0 1 1 0 1 1 0

0 1 1 1 0 0 0

1 0 0 0 1 1 1

1 0 0 1 0 0 1

1 0 1 0 0 1 0

1 0 1 1 1 0 0

1 1 0 0 1 0 0

1 1 0 1 0 1 0

1 1 1 0 0 0 1

1 1 1 1 1 1 1

v uG

v

=

=

c. Answers will vary, but if we choose the 6th row and the 10th row: 0101101 1001001 1102102 1100100 (13th row)

d. v uG

VH uGH

==

0 0 0

0 0 0

0 0 0

0 0 0

GH =

e.

1 1 1

0 1 1

1 0 1

[0 1 0 1 0 0 0] 1 1 0

1 0 0

0 1 0

0 0 1

[1 0 1]

rH =

=

error code: 0010 000 r : 0101 000 0111 000 This is in the codeword list.

Chapter 11 Projects

1193


Project III

a. 1 1 1 1 1 1 1

3 4 5 6 7 8 9TA =

b. 1( )

2.357

2.0357

T TB A A A Y

B

−=−

=

c. 2.0357 2.357y x= −

d. 2.0357 2.357y x= −

Project IV

Answers will vary.

chapter 11 systems of equations and inequalities...the solution of the system is 11, 36 xy== or −...

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