The previous example lent itself to the elimination method because the
coefficients
of
y
were readily eliminated through addition.
But what if this is not the
case and the coefficients of x or y are neither the same nor opposites?
Carefully study
the next example.
Example
Solve the following system by the
elimination strategy
.
x
–
2y = 7
3x + 4y = 1
Neither addition nor subtraction of these two equations will eliminate one of the
variables.
But recall from earlier that we can
multiply
(or
divide
) both sides of an
equation by the same non-zero number
without
changing the solution.
So if we multiply
both sides of the
first
equation by
3
, then
subtract
, we can eliminate
x.
Or
, multiply
both sides of the first equation by
2
then
add
to eliminate
y
.
Both solutions follow.

10C
–
C5-6
–
LG
–
Systems of Linear Equations
Page 21 of 42
Method 1
We will multiply the
first
equation by
3
then
subtract
to eliminate
x
.
3(x
–
2y) = 3(7)
first equation is multiplied by 3 on both sides
3x
–
6y = 21
–
(3x + 4y = 1)
0x
–
10y = 20
-10y = 20
y = -2
Substitute y = -
2 into either original equation.
Let‟s use the first equation.
x
–
2y = 7
x
–
2(-2) = 7
x + 4 = 7
x = 3
The solution is (3, -2)
Method 2
Let‟s multiply the
first
equation by
2
then
add
to eliminate
y
.
2(x
–
2y) = 2(7)
first equation is multiplied by 2 on both sides
2x
–
4y = 14
3x + 4y = 1
5x + 0y = 15
5x = 15
x = 3
Substitute into the first equation to solve for
y
:
x
–
2y = 7
(3)
–
2y = 7
-2y = 4
y = -2
The solution is (3, -2).
As you can see, both methods result in the same solution.
Check the solution to this
system on your own.
Line up like terms
vertically then subtract.
Watch your signs!
Line up like terms
vertically then add.
When using the elimination method of solving linear systems, if the
signs of the coefficients of the variable to be eliminated are the same,
the equations are subtracted.
If the signs of the coefficients of the
variable to be eliminated are opposite, the equations are added.

10C
–
C5-6
–
LG
–
Systems of Linear Equations
Page 22 of 42
In the previous example, we were required to multiply only one equation by a number in
order to eliminate one variable.
This is not always the case, however.
Carefully read
the next example.
Example
Solve the following system using the
elimination strategy
.
2x + 3y = 18
5x
–
4y = -1
If we choose to eliminate
x
, the coefficient of x will need to be
10
in each equation.
Why
10, you ask?
Well, 10 is the
least common multiple
for 2 and 5!
So, we must multiply
the
first
equation by
5
and the
second
equation by
2
then
subtract
to eliminate x.
Method 1
2x + 3y = 18
5x
–
4y = -1
5(2x + 3y) = 5(18)
2(5x
–
4y) = 2(-1)
10x + 15y = 90
–
(10x
–
8y = -2)
0x + 23y = 92
23y = 92
y = 4
Substitute into the second equation to solve for x:
5x
–
4y = -1
5x
–
4(4) = -1
5x
–
16 = -1
5x = 15
x = 3
The solution is (3, 4).
To eliminate
y
, the coefficient of y will need to be
12
in the first equation and -12 in the
second equation.
So we must multiply the
first
equation by
4
on both sides and the
second
equation by
3
on both sides. We then
add
the equations to eliminate y.

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- Winter '19
- jane smith
- Math, Linear Equations, Equations, Systems Of Linear Equations, Elementary algebra