Chapter 4 Section 1 Copyright © 2011 Pearson Education, Inc.

Copyright © 2011 Pearson Education, Inc.

Chapter 4Chapter 4Section 1Section 1

Copyright © 2011 Pearson Education, Inc.

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Systems of Linear Equations in Two Variables

Decide whether an ordered pair is a solution of a linear system.Solve linear systems by graphing.Solve linear systems (with two equations and two variables) by substitution.Solve linear systems (with two equations and two variables) by elimination.Solve special systems.Recognize how a graphing calculator is used to solve a linear system.

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Copyright © 2011 Pearson Education, Inc. Slide 4.1- 3

A set of equations is called a system of equations.

The solution set of a linear system of equations contains all ordered pairs that satisfy all the equations of the system at the same time.


1Objective

Decide whether an ordered pair is a solution of a linear system.


EXAMPLE 1

Is the ordered pair a solution of the given system?

a. (–4, 2)

b. (3, –12)

Replace x with –4 and y with 2 in each equation of the system.

2 6

3 2

x y

x y

2 6x y

(4 )3 22 2( 64)2

28 6 6 6

64 2 2 2

3 2x y

True True

Since (–4, 2) makes both equations true, it is a solution.


continued

b. (3, –12)

Replace x with 3 and y with –12.

2 6

3 2

x y

x y

2 6x y ( 13 ) 23 2 2 63) ( )( 12

6 612 6 6

3 236 33 2

3 2x y

TrueFalse

The ordered pair (3, –12) is not a solution of the system, since it does not make both equations true.


Objective 2

Solve linear systems by graphing.


EXAMPLE 2

Solve the system of equations by graphing.

Graph each linear equation.

The graph suggests that the point of intersection is the ordered pair (–3, 1).

Check the solution in both equations.

2 5

3 6

x y

x y

5

2 5

2

y x

x y

6

23

3

1y

x y

x

Common solution (3, 1)


Objective 3

Solve linear systems (with two equations and two variables) by substitution.


It can be difficult to read exact coordinates, especially if they are not integers, from a graph. For this reason, we usually use algebraic methods to solve systems.

The substitution method, is most useful for solving linear systems in which one equation is solved or can easily be solved for one variable in terms of the other.


EXAMPLE 3

Solve the system.

Since equation (2) is solved for x, substitute 2 – y for x in equation (1).

5 3 6 (1)

2 (2)

x y

x y

5 3 6 (1)yx

5(2 ) 3 6yy

10 5 3 6y y

10 8 6y

8 16y

2y

Be sure to use

parentheses here.

Distributive property

Combine like terms.

Subtract 10.

Divide by –8.


continued

We found y. Now find x by substituting 2 for y in equation (2).

5 3 6 (1)

2 (2)

x y

x y

2x y 22 0

Thus x = 0 and y = 2, giving the ordered pair (0, 2). Check this solution in both equations of the original system.

Check: 5 3 6 (1)yx 5 3(2 6(0) )

0 6 6 6 6

2 (2)yx 0 22 0 0

True

True

The solution set is (0, 2).


Solving a Linear System by Substitution

Step 1 Solve one of the equations for either variable. If one of the equations has a variable term with coefficient 1 or –1, choose that equation since the substitution method is usually easier this way.

Step 2 Substitute for that variable in the other equation. The result should be an equation with just one variable.

Step 3 Solve the equation from Step 2.

Step 4 Find the other value. Substitute the result from Step 3 into the equation from Step 1 to find the value of the other variable.

Step 5 Check the solution in both of the original equations. Then write the solution set.


EXAMPLE 4

Solve the system

Step 1 Solve one of the equations for either x or y.

Step 2 Substitute 5 – 4x into equation (2).

Step 3 Solve.

4 5 (1)

2 3 13. (2)

x y

x y

4 5 (1)x y Subtract 4x.5 4y x

52 3( ) 134x x 2 15 12 13x x

14 28x 2x


continued

Step 4 Now find y.

Step 5 Check the solution (2, –3) in both equations.

4 5 (1)

2 3 13 (2)

x y

x y

5 4y x 5 4( 3)2y

4 5 (1)yx 2 3 13 (2)yx ( )2)4 5( 3 8 3 5

5 5

(2 3) 13(2 3) 4 9 13

13 13

The solution set is (2, –3).

True True


EXAMPLE 5Solve the system

Clear the fractions in equation (2). Multiply by the LCD, 4.

Solve equation (3) for y.

2 5 22 (1)

1 1 1. (2)

2 4 2

x y

x y

1 1 1

24 4

4 2x y

1 1 1

2 44 4

24 x y

2 2 (3)x y

2 2 (3)x y 2 2y x


continuedSubstitute y = 2 – 2x for y in equation (1).

Solve y.

2 5 22x y

2 5 22 (1)

1 1 1 (2)

2 4 2

x y

x y

2 5( 22 )2 2x x 2 10 10 22x x

12 10 22x 12 12x

1x

2 2y x 2 2 )1(y

42 2y

A check verifies that the solution set is {(– 1, 4)}.


Objective 4

Solve linear systems (with two equations and two variables) by elimination.


EXAMPLE 5

Solve the system

Adding the equations together will eliminate x.

To find x, substitute –1 for y in either equation.

2 3 10 (1)

2 2 5. (2)

x y

x y

2 3 10 (1)

2 2 5 (2)

x y

x y

5 5y 1y

2 2 5 (2)x y )12 2( 5x

2 2 5x 2 7x

7

2x

The solution set is {(7/2, –1)}.


Solving a Linear System by Elimination

Step 1 Write both equations in standard form Ax + By = C.

Step 2 Make the coefficients of one pair of variable terms opposites. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either the x- or y-terms is 0.

Step 3 Add the new equations to eliminate a variable. The sum should be an equation with just one variable.

Step 4 Solve the equation from Step 3 for the remaining variable.

Step 5 Find the other value. Substitute the result of Step 4 into either of the original equations and solve for the other variable.

Step 6 Check the solution in both of the original equations. Then write the solution set.


EXAMPLE 7

Solve the system

Step 1 Both equations are in standard form.

Step 2 Select a variable to eliminate, say y. Multiply equation (1) by 7 and equation (2) by 3.

Step 3 Add.

Step 4 Solve for x.

2 3 19 (1)

3 7 6. (2)

x y

x y

14 21 133

9 21 18

x y

x y

23 115x 5x


continued

Step 5 To find y substitute 5 for x in either equation (1) or equation (2).

Step 6 The solution is (5, 3). To check substitute 5 for x and 3 for y in both equations (1) and (2).

The ordered pair checks, the solution set is {(5, 3)}.

2 3 19 (1)

3 7 6 (2)

x y

x y

2 3 19 (1)yx 2 35) 1( 9y 10 3 19y

3 9y 3y


Objective 5

Solve special systems.


EXAMPLE 8

Solve the system

Multiply equation (1) by 4 and add the result to equation (2).

Equations (1) and (2) are equivalent and have the same graph. The equations are dependent.The solution set is the set of all points on the line with equation 2x + y = 6, written in set-builder notation {(x, y) | 2x + y = 6}.

2 6 (1)

8 4 24. (2)

x y

x y

8 4 24 (1)

8 4 24 (2)

x y

x y

0 0 True


EXAMPLE 9

Solve the system

Multiply equation (1) by 2 and add the result to equation (2).

The result of adding the equations is a false statement, which indicates the system is inconsistent. The graphs would be parallel lines. There are no ordered pairs that satisfy both equations. The solution set is .

4 3 8 (1)

8 6 14. (2)

x y

x y

0 2 False

8 6 16 (1)

8 6 14 (2)

x y

x y


Special Cases of Linear Systems

If both variables are eliminated when a system of linear equations is solved,

1. there are infinitely many solutions if the resulting statement is true;

2. there is no solution if the resulting statement is false.


Objective 6

Recognize how a graphing calculator is used to solve a linear system.


EXAMPLE 10

Use a graphing calculator to solve the system

Solve each equation for y.

The points of intersection are displayed at the bottom of the screen, indicating that the solution set is (–3, 1).

2 5

3 6.

x y

x y

5 2y x 1

23

y x

Chapter 4 Section 1 Copyright © 2011 Pearson Education, Inc.

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