The Elimination – by – Addition Method
In Section IV.A, we considered the substitution method of solving systems of two linear equations with two variables. As we mentioned, this method is especially appropriate when it is easy to solve one of the equations for one of the variables without using of fractions. If it is not possible, other methods may be more convenient to use, in particular, the elimination method that we consider here.
Let’s start with the system of equations with two variables in general form: ax
 dx

 by ey

 c f
.
Note
. Further, we will often use the phrase: “multiply an equation by some value”. We mean by that the multiplication of both sides of the equation (or all of its terms) by this value.
The elimination method suggests to eliminate (get rid of) one of the variables in order to get an equation with only one (another) variable. So, the first step in using of this method is to make a decision which variable to eliminate: x or y . Theoretically, it does not matter.
Suppose we decided to eliminate x . The first equation contains term ax , and the second – term dx . If we multiply the first equation by d , and the second equation by a , these terms become equal. Therefore, if (after multiplication) we subtract the first and second equations (in any order), the variable x will be canceled (eliminated). Technically, it would be more convenient to multiply the first equation by d , the second one by – a , and then add the equations (instead of subtraction). Symbolically, we can mark this choice of multiplication like this d
– a ax
 dx

 by
 ey
 c f
.
Note . To eliminate x , we could also multiply the first equation by – d , the second one – by a , and again add them up. If we would decide to eliminate y , we would use coefficients b and e in a similar manner as factors for equations.

The Elimination – by – Addition Method
Example 1 . Solve the following system by the elimination method.
2

5 x x


3 y
2 y

19

20
Solution . First of all we need to make a decision which variable to eliminate. As we mentioned, theoretically it does not matter, so let’s eliminate x by multiplying the first equation by 5, and the second equation – by –2. We represent this decision like this
5
– 2
2

5 x x


3 y
2 y

19

20
After multiplication, the system takes the form
10 x

10 x

15 y

4 y


95

40
.
Now we add these equations, and variable x is eliminated: 15 y – 4 y = 95 – 40. From here,
11 y = 55, y = 5. To find x , we substitute y = 5 in either of the original equations. Let’s use the first one: 2 x

3

5

19 , 2 x + 15 = 19, 2 x = 19 – 15 = 4, x = 2.
Final answer: x = 2, y = 5, or as a solution set, {(2, 5)}.

The Elimination – by – Addition Method
Example 2 . Solve the following system by the elimination method.
4

9 x x


7
2 y y


1
2
Solution . Again, it does not matter which variable to eliminate. Notice that coefficients for y in these equations have opposite signs: positive 7 and negative 2. In this case it’s a bit more convenient to eliminate y by multiplying the first equation by 2, and the second one – by 7:
2
7
4

9 x x


7
2 y y


1
2
.
After multiplication, the system takes the form
8 x

63 x


14 y
14 y


4
.
7
We add these equations to eliminate y : 8 x + 63 x = 4 + 7. From here, 71 x = 11, x

11
71
.
To find y
, let’s use the first equation of the original system: 4 x + 7 y = 2. Substitute here x

11
71
. We will get 4

11
71
7 y


7
2 y

2 ,

44
71

44

7
71
2

71

71 y

2 ,
44

98
71
, y

98
71

7

98
71

1
7

14
71
.
Final answer: x

11
71
, y

14
71
, or as a solution set,
11
71
,
14
71



.