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Chapter Chapter 44Section Section 33
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solving Systems of Linear Equations by Elimination
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33
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4.34.34.34.3Solve linear systems by elimination.Multiply when using the elimination method. Use an alternative method to find the second value in a solution.Use the elimination method to solve special systems.
44
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 11
Slide 4.3 - 3
Solve linear systems by elimination.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 4.3 - 4
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, 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.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1 Using the Elimination Method
Solution:
Slide 4.3 - 5
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
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solving a Linear System by Elimination
Slide 4.3 - 6
In general, use the following steps to solve a linear system of equations by the elimination method.
Step 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.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2 Using the Elimination Method
Solution:
Slide 4.3 - 7
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
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 22
Multiply when using the elimination method.
Slide 4.3 - 8
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3Multiplying Both Equations When Using the Elimination Method
Solution:
Slide 4.3 - 9
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
When using the elimination method, remember to multiply both sides of an equation by the same nonzero number.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 33
Use an alternative method to find the second value in a solution.
Slide 4.3 - 10
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Finding the Second Value by Using an Alternative Method
Solution:
Slide 4.3 - 11
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
The solution set is
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
42
16, .
11
26
42
13y
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 44
Use the elimination method to solve special systems.
Slide 4.3 - 12
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5Using the Elimination Method for an Inconsistent System or Dependent Equations
Solution:
Slide 4.3 - 13
Solve each system by the elimination method.3 7
6 2 5
x y
x y
.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
