Solve systems of equations using the addition/elimination method
.
When solving systems we have found that graphing is very limited when solving
equations. We then considered a second method known as substituion. This is
probably the most used idea in solving systems in various areas of algebra.
However, substitution can get ugly if we don’t have a lone variable. This leads us
to our second method for solving systems of equations. This method is known as
either Elimination or Addition.
We will set up the process in the following
examples, then define the five step process we can use to solve by elimination.
Example 1.
3
x

4
y
=8
5
x
+4
y
=

24
Notice opposites in front of
y
’
s
.
Add columns
.
8
x
=

16
Solve for
x,
divide by
8
8
8
x
=

2
We have our
x
!
5(

2)+4
y
=

24
Plug into either original equation
,
simplify

10
+4
y
=

24
Add 10 to both sides
+
10
+
10
4
y
=

14
Divide by
4
4
4
y
=

7
2
Now we have our
y
!
parenleftbigg

2
,

7
2
parenrightbigg
Our Solution
In the previous example one variable had opposites in front of it,

4
y
and
4
y
.
Adding these together eliminated the
y
completely. This allowed us to solve for
the
x
. This is the idea behind the addition method. However, generally we won’t
have opposites in front of one of the variables. In this case we will manipulate the
equations to get the opposites we want by multiplying one or both equations (on
both sides!). This is shown in the next example.
Example 2.

6
x
+5
y
=
22
2
x
+3
y
=2
We can get opposites in front of
x,
by multiplying the
second equation by
3
,
to get

6
x
and
+6
x
3(2
x
+3
y
)=(2)3
Distribute to get new second equation
.
Page 142
6
x
+9
y
=6
New second equation

6
x
+5
y
=
22
First equation still the same
,
add
14
y
=
28
Divide both sides by 14
14
14
y
=2
We have our
y
!
2
x
+3(2)=2
Plug into one of the original equations
,
simplify
2
x
+6=2
Subtract
6
from both sides

6

6
2
x
=

4
Divide both sides by
2
2
2
x
=

2
We also have our
x
!
(

2
,
2)
Our Solution
When we looked at the
x
terms,

6
x
and
2
x
we decided to multiply the
2
x
by 3
to get the opposites we were looking for.
What we are looking for with our
opposites is the least common multiple (LCM) of the coefficients. We also could
have solved the above problem by looking at the terms with
y
,
5
y
and
3
y
. The
LCM of 3 and 5 is 15. So we would want to multiply both equations, the
5
y
by 3,
and the
3
y
by

5
to get opposites, 15
y
and

15
y
. This illustrates an important
point, some problems we will have to multiply both equations by a constant (on
both sides) to get the opposites we want.
Example 3.
3
x
+6
y
=

9
2
x
+9
y
=

26
We can get opposites in front of
x,
find LCM of
6
and
9
,
The LCM is 18
.
We will multiply to get 18
y
and

18
y
3(3
x
+6
y
)=(

9)3
Multiply the first equation by
3
,
both sides
!
9
x
+
18
y
=

27

2(2
x
+9
y
)=(

26
)(

2)
Multiply the second equation by

2
,
both sides
!

4
x

18
y
=
52
9
x
+
18
y
=

27
Add two new equations together

4
x

18
y
=
52
5
x
=
25
Divide both sides by
5
5
5
x
=5
We have our solution for
x
Page 143
3(5)+6
y
=

9
Plug into either original equation
,
simplify
15
+6
y
=

9
Subtract 15 from both sides

15

15
6
y
=

24
Divide both sides by
6
6
6
y
=

4
Now we have our solution for
y
(5
,

4)
Our Solution
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 Summer '19
 Addition, Subtraction, Elementary arithmetic, Negative and nonnegative numbers