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12
Systems of Linear
Equations and Inequalities
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 1
12.3 Solving Systems of Linear Equations by
Elimination
Objectives
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 © 2014, 2010, 2007 Pearson Education, Inc.
Slide 2
Solve Linear Systems by Elimination
Recall if A = B and A + C = B + C.
This addition can be taken a step further. Adding equal
quantities, rather than the same quantity, to both sides 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. When using this method, the idea is to
eliminate one of the variables. To do this, one of the variables
in the two equations must have coefficients that are opposites.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 3
Solve Linear Systems by Elimination
Example 1
Use the elimination method to solve the system.
2x – y = 5
+ x+y=7
3x
3
= 12
3
x=4
x+y=7
4+y=7
–4
–4
y=3
Each equation in this system is a statement of
equality, so the sum of the left sides equals the
sum of the right sides.
Notice that y has been eliminated. Now we can
solve the result for x.
To find the y-value of the solution, substitute 4
for x into either equation.
The solution set of the system is {(4, 3)}.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 4
Solve Linear Systems by Elimination
CAUTION
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 © 2014, 2010, 2007 Pearson Education, Inc.
Slide 5
Solve Linear Systems by Elimination
Solving a Linear System by Elimination
Step 1 Write both equations in standard form Ax + By = C.
Step 2 Transform 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-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 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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 6
Multiply When Using the Elimination Method
Example 3
Solve the system.
3x + y = 8
5x – 2y = 6
2(3x + y)= (8)2
5x – 2y = 6
Step 1 Both equations are
already in standard form.
6x + 2y = 16
+ 5x – 2y = 6
Step 2 The y-terms have
opposite signs, but their
coefficients are different.
However, if we multiply both
sides of the first equation by
2, then the y terms will cancel
when we add.
11x
11
= 22
11
x=2
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Step 3 Now
when we add,
the y terms are
eliminated.
Step 4 We are
left with an
equation with
just x, which
we can solve
easily.
Slide 7
Multiply When Using the Elimination Method
Example 3 (concluded)
Solve the system.
3x + y = 8
5x – 2y = 6
Step 5 Substitute x = 2 into
either of the equations.
3(2) + y = 8
Step 6 Check that (2,2)
satisfies both equations.
6+y=8
–6
–6
y=2
3(2) + 2 = 8 ?
6+2=8 ?
8=8
5(2) – 2(2) = 6 ?
10 – 4 = 6 ?
The solution set is {(2, 2)}.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
6=6
Slide 8
Use an Alternative Method to Find a Second Value
in a Solution
Example 4
Solve the system.
x + 4y = 9
3x = –2y + 5
–3( x + 4y) =(9)·–3
3x + 2y = 5
We multiply the first
equation by –3 to
eliminate the x’s.
–3x – 12y = –27
+ 3x + 2y = 5
–10y = –22
10 10
11
y
5
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 9
Use an Alternative Method to Find a Second Value
in a Solution
Example 4 (concluded)
Solve the system.
x + 4y = 9
–2(3x + 2y)= (5)·–2
x + 4y = 9
+ –6x – 4y = –10
–5x = –1
5 5
1
x
5
11
5
Substituting –11/5 for y into one of the
given equations would give x, but the
arithmetic involved would be messy.
Instead, solve for x by starting again with
the original equations and eliminating y.
y
 1 11  
The solution set is  ,    .
 5 5  
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 10
Solve Special Systems by Elimination
2
x  y 1
3
Solve the system by the elimination method. 2 x  3 y  3
Multiply each side of the first equation by 3; then add the two
equations.
Example 5
3( ⅔x – y) = ( 1)3
–2x + 3y = –3
2x – 3y = 3
+ –2x + 3y = –3
0=0
True.
A true statement occurs when the equations are equivalent.
As before, this indicates that every solution of one equation
is also a solution of the other. The solution set is
 x, y  2 x  3 y  3 .
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 11
Solve Special Systems by Elimination
Example 5 (concluded)
x  4y  3
Solve the system by the elimination method. 2 x  8 y  5
–2(x + 4y)=(3)·–2
2x + 8y = 5
–2x – 8y = –6
+ 2x + 8y = 5
0 = –1
False.
The false statement 0 = –1 indicates that the solution set
is 0 .
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 12