Solving Systems of Linear Equations by Elimination

Solve linear systems by elimination.

Multiply when using the elimination method.

Use an alternative method to find the second value in a solution.

Solve special systems by elimination.

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Copyright © 2012 Pearson Education, Inc.

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Objective 1

Solve linear systems by elimination.

Slide 4.3-3

An algebraic method that depends on the addition property of equality can also be used to solve systems. Adding the same quantity to each side of an equation results in equal sums:

If A = B, then A + C = B + C.

Solve linear systems by elimination.

We can take this addition a step further. Adding equal quantities, rather than the same quantity, to each side of an equation also results in equal sums:

If A = B and C = D, then A + C = B + D.

Using the addition property to solve systems is called the elimination method. With this method, the idea is to eliminate one of the variables. To do this, one pair of variable terms in the two equations must have coefficients that are opposite.

Slide 4.3-4

Solution:

Solve the system.

3 7

2 3

x y

x y

2 33 7x yx y 2 2 3y 5 0

5 5

1x

2x

4 44 3y

1y

2, .1The solution set is

A system is not completely solved until values for both x and y are found. Do not stop after finding the value of only one variable. Remember to write the solution set as a set containing an ordered pair

Slide 4.3-5

EXAMPLE 1 Using the Elimination Method

Solving a Linear System by Elimination

Solving a Linear System by EliminationStep 1: Write both equations in standard form, Ax + By = C.

Step 2: Transform the equations as needed so that the coefficients of one pair of variable terms are opposites. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either the x- or y-term is 0.

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

Step 5: Substitute the result from Step 4 into either of the original equations, and solve for the other variable.

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

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

* It does not matter which variable is eliminated first. Choose the one that is more convenient to work with.

Slide 4.3-6

Solution:

Solve the system.

2

2 10

x y

x y

2 22x yy y 2x y

2 02 1y xx y 3 2

3 3

1x

4, .2The solution set is

2 10x yy y 2 10x y

4x

4 44 2y 2y

Slide 4.3-7

EXAMPLE 2 Using the Elimination Method

Objective 2

Multiply when using the elimination method.

Slide 4.3-8

Sometimes we need to multiply each side of one or both equations in a system by some number before adding will eliminate a variable.

Multiply when using the elimination method.

Slide 4.3-9

When using the elimination method, remember to multiply both sides of an equation by the same nonzero number.

Solution:

Solve the system.

4 5 18

3 2 2

x y

x y

4 18 252 x y 8 10 36x y

15 10 110 36 08 xx yy 2

23 23

3 46x

.2, 2The solution set is

3 25 52x y

15 10 10x y

2x

66 62 2y 2 4

2 2

y

3 22 2y

2y

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EXAMPLE 3 Using the Elimination Method

Objective 3

Use an alternative method to find the second value in a solution.

Slide 4.3-11

Sometimes it is easier to find the value of the second variable in a solution by using the elimination method twice.

Use an alternative method to find the second value in a solution.

Slide 4.3-12

When using the elimination method, remember to multiply both sides of an equation by the same nonzero number.

Solution:

Solve the system.3 8 4

6 9 2

y x

x y

4 32 28x y 6 23 9 3x y

8 6 16x y 18 6 27x y

11

4 33 38x y

26x

12 9 24x y 12 4 18x y

6 2 292 x y

13 42y +

11

26x

26

2 1

26

6 1x

+

13 1

3

3

1 42y

The solution set is42

16, .

11

26

42

13y

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EXAMPLE 4 Finding the Second Value by Using an Alternative Method

Objective 4

Solve special systems by elimination.

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3 7

6 2 5

x y

x y

Solution:

Solve each system by the elimination method.

.The solution set is

2 5 1

4 10 2

x y

x y

72 3 2x y 6 2 5x y

6 2 14x y 6 2 5x y +

0 19

4 10 2x y 4 10 2x y +

0 0

2 5 212 x y 4 10 2x y

, 2 5 1 .x y x y The solution set is

Slide 4.3-15

EXAMPLE 5 Solving Special Systems Using the Elimination Method