3.3 Solving Linear Systems by
Linear Combination
10/12/12
Solving Systems of Equations
So far, we have solved systems using
graphing and substitution.
 The third method is called Linear
Combination or Elimination.

Linear Combination or
Elimination
Is a method of solving a system of equations
by multiplying one or both equations by a
constant, then adding the revised equations
to eliminate a variable
Linear Combination Method
Step 1: Multiply one or both of the equations
by a constant, if necessary, to obtain
coefficients that differ only in sign for one of
its variables.
Step 2: Add the revised equations from Step 1.
Combining like terms will eliminate one variable.
Solve for the remaining variable.
Step 3: Substitute the value obtained in Step 2
into either of the original equations and solve
for the other variable.
Step 4: Check the solution in each of the
original equations.
Solve
x  2y  4
4x  2 y  6
Method: by Graphing
1
y x2
2
y  2 x  3
Solution (2, -1)
Solve
x  2y  4
4x  2 y  6
Method: Substitution
x  2y  4
4( 2 y  4)  2 y  6
x  2(1)  4
x  2  4
x2
8 y  16  2 y  6
10 y  10
y  1
Solution (2, -1)
Solve
x  2y  4
4x  2 y  6
x  2y  4
+ 4x  2 y  6
5x = 10
5
5
x=2
2 - 2y = 4
-2
-2
-2y = 2
-2 -2
Method: Linear
Combination
Step 1: Multiply one or both of the
equations by a constant, if necessary, to
obtain coefficients that differ only in
sign for one of its variables.
Step 2: Add the revised equations from
Step 1. Combining like terms will
eliminate one variable. Solve for the
remaining variable.
Step 3: Substitute the value obtained in
Step 2 into either of the original
equations and solve for the other
variable.
y = -1
Solution (2, -1)
Example 1
Multiply One Equation
Solve the linear system using the linear combination
method.
2x – 3y = 6
Equation 1
4x – 5y = 8
Equation 2
SOLUTION
STEP 1 Multiply the first equation by – 2 so that the
coefficients of x differ only in sign.
2x – 3y = 6
– 4x + 6y = –12
4x – 5y = 8
4x – 5y = 8
STEP 2 Add the revised equations
y = –4
and solve for y.
Example 1
Multiply One Equation
STEP 3 Substitute – 4 for y in one of the original
equations and solve for x.
2x – 3y = 6
2x – 3( – 4) = 6
2x + 12 = 6
2x = – 6
x = –3
Write Equation 1.
Substitute – 4 for y.
Simplify.
Subtract 12 from each side.
Solve for x.
STEP 4 Check by substituting – 3 for x and – 4 for y in
the original equations.
ANSWER
The solution is ( – 3, – 4 ).
Example 2
Multiply Both Equations
Solve the system using the linear combination method.
7x – 12y = – 22
Equation 1
– 5x + 8y = 14
Equation 2
SOLUTION
STEP 1 Multiply the first equation by 2 and the second
equation by 3.
7x – 12y = – 22
– 5x + 8y = 14
STEP 2 Add the revised equations
and solve for x.
14x – 24y = – 44
–15x + 24y = 42
–x
= –2
x = 2
Example 2
Multiply Both Equations
STEP 3 Substitute 2 for x in one of the original equations
and solve for y.
– 5x + 8y = 14
Write Equation 2.
– 5( 2) + 8y = 14
Substitute 2 for x.
–10 + 8y = 14
y =3
Multiply.
Solve for y.
STEP 4 Check by substituting 2 for x and 3 for y in the
original equations.
ANSWER
The solution is (2, 3).
Checkpoint
Solve a Linear System
Solve the system using the linear combination method.
1. x – 4y = 5
ANSWER
(1, – 1)
2x + y = 1
2. 3x – 2y = 2
4x – 3y = 1
ANSWER
(4, 5 )
Homework:
3.3 p.142 #8-18