Transcript Solving Systems of Linear Equations by Elimination

4.3 Solving Systems of Linear Equations by Elimination
1
Solve linear systems by elimination.
2
Multiply when using the elimination method.
3
Use an alternative method to find the second value in a solution.
4
Solve special systems by elimination.
Copyright © 2012 Pearson Education, Inc.
Objective 1
Solve linear systems by
elimination.
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 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
EXAMPLE 1 Using the Elimination Method
Solve the system.
3x  y  7
2x  y  3
Solution:
 3x  y    2 x  y   7  3
2  2  y  3
5 x 10

5
5
x2
The solution set is  2, 1.
4 y  4  3 4
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
Slide 4.3-5
Solving a Linear System by Elimination
Solving a Linear System by Elimination
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.
Slide 4.3-6
EXAMPLE 2 Using the Elimination Method
Solve the system.
x  2  y
2 x  y  10
Solution:
x  2 y  2  y  y  2
x y 2
2 x  y  y  10  y
2 x  y  10
The solution set is
 4, 2.
 x  y    2x  y   2  10
3 x 12

3
3
x4
4 y4  24
y  2
Slide 4.3-7
Objective 2
Multiply when using the
elimination method.
Slide 4.3-8
Multiply when using the elimination method.
Sometimes we need to multiply each side of one or both equations in a
system by some number before adding will eliminate a variable.
When using the elimination method, remember to multiply both sides of an
equation by the same nonzero number.
Slide 4.3-9
EXAMPLE 3 Using the Elimination Method
Solve the system.
4 x  5 y  18
3x  2 y  2
Solution:
 2 4x  5 y   18  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
Slide 4.3-10
Objective 3
Use an alternative method to find
the second value in a solution.
Slide 4.3-11
Use an alternative method to find the second value in
a solution.
Sometimes it is easier to find the value of the second variable in a
solution by using the elimination method twice.
When using the elimination method, remember to multiply both sides of an
equation by the same nonzero number.
Slide 4.3-12
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
8 x  6 y  16
+ 18 x  6 y  27
26x
 11
2 6 x 1 1

26
26
11
x
26
3 4x  3 y   8 3
 2 6 x  2 y  9  2
12 x  9 y  24
+ 12 x  4 y  18
13 y  42
13 y 42

13 13
42
y
13
 11 42  
The solution set is  ,   .
 26 16  
Slide 4.3-13
Objective 4
Solve special systems by
elimination.
Slide 4.3-14
EXAMPLE 5 Solving Special Systems Using the Elimination Method
Solve each system by the elimination method.
3x  y  7
6x  2 y  5
2x  5 y  1
4 x  10 y  2
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
.
 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.
Slide 4.3-15