chapter 5 detailed summary

Detailed Summary 1

CHAPTER 5 DETAILED SUMMARY

SECTION 5.1 Systems of Linear Equations in Two Variables and Applications

KEY CONCEPTS EXAMPLES

Two or more linear equations taken together form a system of linear equations. A solution to a system of equations in two variables is an ordered pair that is a solution to each individual equation. Graphically, this is a point of intersection of the graphs of the equations.

The substitution method and the addition method are often used to solve a system of linear equations in two variables.

Example 1:

Solve by using the substitution method.

x 1 5y 5 21 x 5 21 2 5y2x 1 3y 5 5

2(21 2 5y) 1 3y 5 5 Substitute 21 2 5y for x. 22 2 10y 1 3y 5 5

27y 5 7 y 5 21

Back substitution: x 5 21 2 5y5 21 2 5(21)5 4

The solution set is {(4, 21)}.

A system of linear equations in two variables will have no solution if the equations in the system represent parallel lines. In such a case, we say that the system is inconsistent.

Example 2:

Solve by using the addition method.

2x 2 y 5 4 2x 2 y 5 46(x 2 1) 5 3y 6x 2 3y 5 6

2x 2 y 5 4 26x 1 3y 5 2126x 2 3y 5 6 6x 2 3y 5 6

0 5 26This results in a contradiction. There is no solution to the system, and the system is inconsistent. The solution set is { }.

A system of linear equations in two variables will have infi nitely many solutions if the equations represent the same line. In such a case, we say that the equations are dependent. The solution set to the system is the set of points on the line.

Example 3:

13x 1 1

3y 5 1 x 1 y 5 32x 5 6 2 2y 2x 1 2y 5 6

x 1 y 5 3 22x 2 2y 5 262x 1 2y 5 6 2x 1 2y 5 6

0 5 0The system results in an identity. The equations represent the same line x 1 y 5 3. Solving for y in terms of x yields y 5 3 2 x. Thus, the solutions to the system are ordered pairs of the form (x, 3 2 x). The solution set can be written as

{(x, 3 2 x) 0 x is any real number}

Likewise, by solving for x in terms of y, the solution set can be written as

{(3 2 y, y) 0 y is any real number}

31 2 4 5 6 7

1

2

3

45

212223

24

25

212223x

y

2x 1 3y 5 5

x 1 5y 5 21

(4, 21)

Solve for x.

Standard form.

Multiply by 23.

1 2 3 4 5

1

2

3

45

212223

24

25

2122232425x

y

6(x 2 1) 5 3y

2x 2 y 5 4

Multiply by 3.

Standard form

Multiply by 22.

1 2 3 4 5 6

1

2

3

45

6

212223

24

21222324x

y

2x 5 6 2 2y x 1 y 5 113

13

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2 Chapter 5 Systems of Equations and Inequalities

SECTION 5.2 Systems of Linear Equations in Three Variables and Applications

KEY CONCEPTS

A linear equation in three variables is an equation that can be written in the form

Ax 1 By 1 Cz 5 Dwhere A, B, and C are not all zero.

A solution to a system of linear equations in three variables is an ordered triple (x, y, z) that satisfi es each equation in the system. Geometrically, a solution is a point of intersection of the planes represented by the equations in the system.

When solving a system of linear equations, if a contradiction arises such as 0 5 4, the system is inconsistent and has no solution.

If an identity arises such as 0 5 0, the equations are dependent and the system has infi nitely many solutions.

EXAMPLES

Example 1:

A 2x 1 4y 1 3z 5 7B 3x 2 z 5 4C y 1 5z 5 23

A 2x 1 4y 1 3z 5 724 ? C 2 4y 2 20z 5 12 2x 2 17z 5 19 D

B 3x 2 z 5 4 26x 1 2z 5 28D 2x 2 17z 5 19 6x 2 51z 5 57 249z 5 49 z 5 21

From equation B we have 3x 2 (21) 5 4 x 5 1

From equation C we have y 1 5(21) 5 23 y 5 2

The ordered triple (1, 2, 21) checks in all three original equations.

The solution set is {(1, 2, 21)}.

Example 2:

A 3x 1 y 5 6B 8x 1 3y 2 z 5 2 8x 1 3y 2 z 5 2C 25x 2 2y 1 z 5 4 25x 2 2y 1 z 5 4 3x 1 y 5 6 D A 3x 1 y 5 621 ? D 23x 2 y 5 26 0 5 0 (Identity)

The equations are dependent. To fi nd the general solution, express the dependency among the variables as an ordered triple. From equation A , we can solve for y in terms of x.

A y 5 6 2 3x

From equation B or C we can substitute y 5 6 2 3x and then solve for z in terms of x.

C 25x 2 2(6 2 3x) 1 z 5 4 x 2 12 1 z 5 4 z 5 16 2 x

The solution set can be written as

{(x, 6 2 3x, 16 2 x) 0 x is any real number}.

Alternatively, the solution set can be written as

e a6 2 y

3, y,

y 1 42

3b ` y is any real number f or

{(16 2 z, 3z 2 42, z) 0 z is any real number}.

Eliminate one variable from the system to create a system of two equations and two variables.

We will eliminate variable y first.

Multiply by 22.

Multiply by 3.


Detailed Summary 3

SECTION 5.3 Partial Fraction Decomposition

KEY CONCEPTS

Partial fraction decomposition is used to write a rational expression as a sum of simpler fractions.

There are two basic parts to fi nd the partial fraction decomposition of a rational expression.

I. Factor the denominator of the expression into linear factors and quadratic factors that are not further factorable over the integers. Then set up the “form” or “structure” for the partial fraction decomposition into simpler fractions.

II. Next, multiply both sides of the equation by the LCD. Then set up a system of linear equations to find the coefficients of the terms in the numerator of each fraction.

Note: The numerator of the original rational expression must be of lesser degree than the denominator. If this is not the case, fi rst use long division.

EXAMPLES

Example 1:

4x3 1 4x2 1 8x 1 12

x4 1 4x2 54x3 1 4x2 1 8x 1 12

x2(x2 1 4)

Set up the format for the decomposition.

4x3 1 4x2 1 8x 1 12

x2(x2 1 4)5

Ax

1B

x2 1Cx 1 D

x2 1 4

Multiply by the LCD to clear fractions.

x2(x2 1 4) c 4x3 1 4x2 1 8x 1 12

x2(x2 1 4)d 5

x2(x2 1 4) c Ax

1B

x2 1Cx 1 D

x2 1 4d

Apply the distributive property and combine like terms.

4x3 1 4x2 1 8x 1 12 5 Ax(x2 1 4) 1 B(x2 1 4) 1 (Cx 1 D)x2

5 Ax3 1 4Ax 1 Bx2 1 4B 1 Cx3 1 Dx2

Therefore,

4x3 1 4x2 1 8x 1 12 5 (A 1 C)x3 1 (B 1 D)x2 1 4Ax 1 4B

Equate the coeffi cients on like terms and solve the resulting system for A, B, C, and D.

4 5 A 1 C4 5 B 1 D

A 5 2, B 5 3, C 5 2, D 5 18 5 4A

12 5 4B

The partial fraction decomposition is:

4x3 1 4x2 1 8x 1 12

x4 1 4x2 52x

13

x2 12x 1 1

x2 1 4



SECTION 5.4 Systems of Nonlinear Equations in Two Variables

KEY CONCEPTS

A nonlinear system of equations is a system in which one or more equations is nonlinear.

The substitution method is often used to solve a nonlinear system of equations.

In some cases, the addition method can be used provided that the terms containing the corresponding variables are like terms.

Example 2:

Use the addition method to solve the system.

A x2 1 2y2 5 11 2x2 2 2y2 5 211 x2 1 3y2 5 12

y2 5 1 y 5 61

B x2 1 3y2 5 12

From equation A , we have

If y 5 1, x2 1 2(1)2 5 11 x2 5 9 x 5 6 3

If y 5 21, x2 1 2(21)2 5 11 x2 5 9 x 5 6 3

The solution set is{(3, 1), (3, 21), (23, 1), (23, 21)}.

EXAMPLES

Example 1:

Use the substitution method to solve the system.

A y 5 2x 2 1 A y 5 2x 2 1

B y 5 x2 1 6x 1 9 B y 5 x2 1 6x 1 9

B 2x 2 1 5 x2 1 6x 1 90 5 x2 1 7x 1 100 5 (x 1 5)(x 1 2)

x 5 25 or x 5 22

From equation A , we have

If x 5 25, y 5 2(25) 2 1 y 5 4 (25, 4)If x 5 22, y 5 2(22) 2 1 y 5 1 (22, 1)

The solution set is {(25, 4), (22, 1)}.

The solutions are the points of intersection of the graphs of the two equations.

Multiply by 21.

24252627 23 1 2 321

22

23

7

3

4

5

6

2

1

2122x

yy 5 x2 1 6x 1 9

y 5 2x 2 1

(22, 1)

(25, 4)


Detailed Summary 5

SECTION 5.5 Inequalities and Systems of Inequalities in Two Variables

KEY CONCEPTS

Linear inequalities in two variables:

An inequality of the form Ax 1 By , C, where A and B are not both zero, is called a linear inequality in two variables. (The symbols ., #, and $ can be used in place of , in the defi nition.)

EXAMPLES

Example 1:x 2 y . 2Graph x 2 y 5 2.

Test a point above or below the line.

1 2 3 4 5

1

2

3

45

212223

24

25

2122232425x

y

Example 2:x2 1 y2 # 4Graph x2 1 y2 5 4.

Test a point inside or outside the circle.

1 2 3 4 5

1

2

3

45

212223

24

25

2122232425x

y

Solving a linear inequality in two variables:

The basic steps to solve a linear inequality in two variables are as follows.

1. Graph the related equation. The resulting line is drawn as a dashed line if the inequality is strict, and is otherwise drawn as a solid line.

2. Select a test point from either side of the line. If the ordered pair makes the original inequality true, then shade the half-plane from which the point was taken. Otherwise, shade the other half-plane.

A nonlinear inequality in two variables is solved using the same basic procedure.

Two or more inequalities in two variables make up a system of inequalities in two variables. The solution set to the system is the region of overlap (intersection) of the solution sets of the individual inequalities.

Example 3:

3x 1 2y , 6y $ 2x 1 3

Graph the related equations and test an ordered pair from a point above or below each line.

3x 1 2y 5 6 y 5 2x 1 3

Test (22, 1) Test (22, 1)

3(22) 1 2(1) ,?

6 Yes 1 $?

2(22) 1 3 Yes

Shade below the line 3x 1 2y 5 6 and above the line y 5 2x 1 3.

Test (0, 0) x 2 y . 2

(0) 2 (0) .?

2 falseShade below the line.

Test (0, 0) x2 1 y2 # 4

(0)2 1 (0)2 #?

4 trueShade inside the circle.

1 2 3 4 5

1

2

3

45

212223

24

25

2122232425x

y

Test point(0, 0)

1 2 3 4 5

1

2

3

45

212223

24

25

2122232425x

y

Test point(0, 0)

1 2 3 4 5

1

2

3

45

212223

24

25

2122232425x

y

y 5 2x 1 3

3x 1 2y 5 6

Test point(22, 1)



SECTION 5.6 Linear Programming

KEY CONCEPTS

A process called linear programming enables us to maximize or minimize a function under specifi ed constraints. The function to be maximized or minimized is called the objective function.

The steps to solve a linear programming application are outlined here.

Step 1 Write an objective function, z 5 f (x, y).

Step 2 Write a system of inequalities defi ning the constraints on x and y.

Step 3 Graph the feasible region and identify the vertices.

Step 4 Evaluate the objective function at each vertex of the feasible region. Use the results to identify the values of x and y that optimize the objective function, and identify the optimal value of z.

EXAMPLES

Example 1:

An artist makes necklaces and earrings. The cost of materials for a necklace is $12 and the cost for a pair of earrings is $6. The artist has time to create at most 40 necklaces per week and at most 60 pairs of earrings per week. Furthermore, the artist does not want to exceed $540 per week in cost of materials. If the profi t on a necklace is $20 and the profi t on a pair of earrings is $15, how many of each item should the artist make to maximize weekly profi t? Assume that all necklaces and earrings made also sell.

Let x represent the number of necklaces.Let y represent the number of pairs of earrings.

Step 1 Maximize profi t: z 5 20x 1 15y

Step 2 x $ 0, y $ 0, x # 40, y # 60

12x 1 6y # 540

Step 3

20 25 30 35 40 45 5050 10 15

40

70

60

50

10

30

20

x

y

(40, 10)

(15, 60)

(40, 0)(0, 0)

(0, 60)

Step 4 Test z 5 20x 1 15y at each vertex:

at (0, 0): z 5 20(0) 1 15(0) 5 $0at (0, 60): z 5 20(0) 1 15(60) 5 $900at (15, 60): z 5 20(15) 1 15(60) 5 $1200at (40, 10): z 5 20(40) 1 15(10) 5 $950at (40, 0): z 5 20(40) 1 15(0) 5 $800

The maximum weekly profi t of $1200 will be obtained if the artist makes 15 necklaces and 60 pairs of earrings.


chapter 5 detailed summary

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