3.2 solving systems algebraically substitution & elimination

3.2 Solving Systems Algebraically

Substitution & Elimination

EXAMPLE 1 Use the substitution method

Solve the system using the substitution method.

2x + 5y = –5

x + 3y = 3

Equation 1

Equation 2

SOLUTION

STEP 1 Solve Equation 2 for x.

x = –3y + 3 Revised Equation 2


STEP 2

Substitute the expression for x into Equation 1 and solve for y.

2x +5y = –5

2( ) + 5y = –5

y = 11

Write Equation 1.

Substitute –3y + 3 for x.

Solve for y.

STEP 3

Substitute the value of y into revised Equation 2 and solve for x.

x = –3y + 3

x = –3( ) + 3

x = –30

Write revised Equation 2.

Substitute 11 for y.

Simplify.

–3y + 3

11


CHECK Check the solution by substituting into the original equations.

2(–30) + 5(11) –5= ? Substitute for x and y. = ? –30 + 3(11) 3

Solution checks. 3 = 3 –5 = –5

The solution is (– 30, 11).

ANSWER

EXAMPLE 2 Use the elimination method

Solve the system using the elimination method.

3x – 7y = 10

6x – 8y = 8

Equation 1

Equation 2SOLUTION

Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign.STEP 1

3x – 7y = 106x – 8y = 8

–6x + 14y = -206x – 8y = 8

STEP 2

Add the revised equations and solve for y. 6y = –12

y = –2


STEP 3

Substitute the value of y into one of the original equations. Solve for x.

3x – 7y = 10

3x – 7(–2) = 10

3x + 14 = 10

x =4

3 – Solve for x.

Simplify.

Substitute –2 for y.

Write Equation 1.


The solution is ( , –2)4

3

–

ANSWER

CHECK

You can check the solution algebraically using the method shown in Example 1. You can also use a graphing calculator to check the solution.

1. 4x + 3y = –2x + 5y = –9

Solve the system using the substitution for 1 and the elimination method for 2.

GUIDED PRACTICE for Examples 1 and 2

The solution is (1,–2).

ANSWER

2. 3x + 3y = –155x – 9y = 3

The solution is ( -3 , –2)

ANSWER

Solve the system using the substitution or the elimination method.

GUIDED PRACTICE for Examples 1 and 2

3. 3x – 6y = 9 –4x + 7y = –16

The solution is (11, 4)

ANSWER

EXAMPLE 4 Solve linear systems with many or no solutions

Solve the linear system.

a. x – 2y = 4

3x – 6y = 8

b. 4x – 10y = 8

– 14x + 35y = – 28

SOLUTION

a. Because the coefficient of x in the first equation is 1, use the substitution method.

Solve the first equation for x.

x – 2y = 4

x = 2y + 4

Write first equation.

Solve for x.


Substitute the expression for x into the second equation.

3x – 6y = 8

3(2y + 4) – 6y = 8

12 = 8

Write second equation.

Substitute 2y + 4 for x.

Simplify.

Because the statement 12 = 8 is never true, there is no solution.

ANSWER


b. Because no coefficient is 1 or – 1, use the elimination method.

Multiply the first equation by 7 and the second equation by 2.

4x – 10y = 8

– 14x + 35y = – 28

28x – 70y = 56

– 28x + 70y = – 56

Add the revised equations. 0 = 0

ANSWER

Because the equation 0 = 0 is always true, there are infinitely many solutions.

GUIDED PRACTICE for Example 4

Solve the linear system using any algebraic method.

5. 12x – 3y = – 9

– 4x + y = 3

SOLUTION

Because the coefficient of y in the second equation is y, use the substitution method.

Solve the 2nd equation for y.

– 4x + y = 3

y = 4x + 3


Solve for y.


Substitute the expression for y into the first equation.

12x – 3y = – 9

12x – 3(4y + 3) = – 9

0 = 0


Substitute 4x + 3 for y.

Simplify.

Because the equation 0 = 0 is always true, there are infinitely many solutions.

ANSWER



6. 6x + 15y = – 12

– 2x – 5y = 9

Because no coefficient is 1 or – 1, use the elimination method. Multiply the second equation by 3

6x + 15y = – 12

– 2x – 5y = 9

Add the revised equations. 0 = 15

ANSWER

Because the statement 0 = 15 is never true, there are no solutions.

6x + 15y = – 12

– 6x – 15y = 273



7. 5x + 3y = 20

– x – y = – 43

5

Because the coefficient of x in the first equation is – 1, use the substitution method.

Solve the 2nd equation for x.


Solve for x.

– x – y = – 43

5

x = – y + 4 3

5


Substitute the expression for x into the first equation.

Write first equation.

Simplify.

Because the statement 20 = 20 is always true, there are infinitely many solution.

ANSWER

5x + 3y = 20

5 ( – y + 4 ) + 3y = 203

520 = 20

Substitute for x.5

3– y + 4



8. 12x – 2y = 21

3x + 12y = – 4

Because no coefficient is 1 or – 1, use the elimination method.

Multiply the second equation by 6

12x – 2y = 21

3x + 12y = – 4

Add the revised equations. 75x = 122

72x – 12y = 126

3x + 12y = – 4

x122

75=

6


Substitute the expression for x into the second equation.

3x + 12y = – 4 Write second equation.

Simplify.

3( ) + 12y = 8122

75 Substitute for x.12275

y = – 37

50

The solution is ( , )

ANSWER

122

75

– 37

50



9. 8x + 9y = 15

5x – 2y = 17

5x + 5y = 510.

5x + 3y = 4.2

(3, –1)

ANSWER

(0.6, 0.4)

ANSWER

3.2 solving systems algebraically substitution & elimination

Documents