Download Solving a Linear System by Elimination

Chapter 4
Section 3
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4.3
1
2
3
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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.
Use the elimination method to solve
special systems.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 1
Solve linear systems by
elimination.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.3 - 3
Solve linear systems by elimination.
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.
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
Slide 4.3 - 4
EXAMPLE 1
Using the Elimination Method
Solve the system.
3x  y  7
2x  y  3
Solution:  3x  y    2 x  y   7  3
5 x 10

5
5
x2
2  2  y  3
4 y  4  3 4
The solution set is  2, 1.
y  1
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
Slide 4.3 - 5
Solving a Linear System by Elimination
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 4: Solve the equation from Step 3 for the remaining variable.
Step 5: Substitute the result from 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.
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
Slide 4.3 - 6
EXAMPLE 2
Solve the system.
Using the Elimination Method
x  2  y
2 x  y  10
Solution:
x  2 y  2  y  y  2
x y 2
 x  y    2x  y   2  10
2 x  y  y  10  y
2 x  y  10
The solution set is  4, 2 .
3 x 12

3
3
x4
4 y4  24
y  2
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Slide 4.3 - 7
Objective 2
Multiply when using the
elimination method.
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Slide 4.3 - 8
EXAMPLE 3
Solve the system.
Solution:
 2 4x  5 y   18  2
Multiplying Both Equations
When Using the Elimination
Method
4 x  5 y  18
3x  2 y  2
8 x  10 y  36
8x 10 y   15x 10 y   36 10
23 x 46

23
23
x  2
The solution set is  2, 2 .
53x  2 y   2 5
15 x  10 y  10
3  2  2 y  2
6  2 y  6  2  6
2y 4

2 2
y2
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
Slide 4.3 - 9
Objective 3
Use an alternative method to
find the second value in a
solution.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.3 - 10
EXAMPLE 4
Finding the Second Value by
Using an Alternative Method
Solve the system.
3y  8  4x
6x  9  2 y
Solution:
 2 4x  3 y   8  2
 3 6x  2 y   9  3
3 4x  3 y   8 3
 2 6 x  2 y  9  2
8 x  6 y  16
+ 18 x  6 y  27
26x
 11
12 x  9 y  24
+ 12 x  4 y  18
13 y  42
26 x 11

26
26
x
13 y 42

13 13
11
26
The solution set is
42
y
13
 11 42  
 ,   .
 26 16  
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Slide 4.3 - 11
Objective 4
Use the elimination method to
solve special systems.
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Slide 4.3 - 12
EXAMPLE 5
Using the Elimination Method
for an Inconsistent System or
Dependent Equations
Solve each system by the elimination method.
3x  y  7
2x  5 y  1
6x  2 y  5
Solution:
 23x  y   7  2
6x  2 y  5
6 x  2 y  14
+ 6x  2 y  5
0  19
The solution set is .
4 x  10 y  2
 2 2x  5 y   1 2
4 x  10 y  2
4 x  10 y  2
+ 4 x  10 y  2
00
The solution set is
 x, y  2 x  5 y  1.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4.3 - 13