Systems of Linear Equations: Elimination by Addition Method
First, let's look at a system of 2 equations with 2 variables:
4 x 5 y= 7
x 2 y= 8
The general idea is to add a "multiple" of one equation to another equation in such a way
as to eliminate one of the variables. Note that a "multiple" of an equation is just a nonzero
constant times an equation.
In the above problem, form the multiple of the second equation that you get by
multiplying the equation by -4. This gives:
4 x 8 y = 32
Of course, be careful with those minus signs!
Now add our multiple to the other equation (the first equation):
4 x 8 y = 32
4 x 5 y= 7
________________
13 y = 39
The new system of equations consists of this new equation and either one of the other
equations. Let's use the following for our equivalent system of equations. (Equivalent
here means that the systems have the same solution.)
4x 5y= 7
13 y = 39
We have now eliminated the x variable in the one equation. Solve for y by dividing both
sides of that final equation by 13. This gives:
y= 3
We then substitute 3 for y in the other equation and solve for x as follows:
4 x 5 ( 3) = 7
4 x 15 = 7
4 x= 8
x= 2
The solution to the original system of equations is thus the point (-2, 3). In other words,
we get x = -2 and y = 3. You can verify that this is indeed a solution by substituting these
values into the original equations.
Next, let's try a harder system of 2 equations with 2 variables:
2x 3y= 1
3 x 4 y= 3
Without using fractions we have no way to add a multiple of one equation to the other in
order to eliminate one of the variables. However, you can also add a multiple of one
equation to a multiple of another equation. That makes things easier in this case since you
can take 3 times the first equation, -2 times the second equation, and then add the
resulting equations. Let's try it. First write down the two multiples just mentioned and
then add them:
6 x 9 y= 3
6 x 8 y= 6
________________
y= 3
The new, equivalent system of equations consists of this new equation and one of the
original equations:
2x 3y= 1
y= 3
We have now eliminated x. The second equation is easy to solve for y:
y=3
Then we substitute 3 for y in the other equation to get:
2 x 3 ( 3) = 1
2x 9=1
2 x = 10
x= 5
Thus the solution to the system is the point (5, 3). You can check the solution by
substituting 5 for x and 3 for y in the original equations.
Now, let's try a system of 3 equations with 3 variables:
x 2y z = 4
2 x 4 y 3 z= 1
3 x 6 y 7 z= 4
The first goal is to eliminate one of the variables from two of the equations. Since it looks
to be easy to eliminate x, let's aim for that. Add -2 times the first equation to the second
as shown below:
2 x 4 y 2 z= 8
2 x 4 y 3 z= 1
___________________
8 y 5 z= 7
Similarly, add 3 times the first equation to the third equation:
3 x 6 y 3 z = 12
3x 6y 7z = 4
___________________
12 y 10 z = 8
The new, equivalent system of equations consists of one of the original equations and the
2 new, shorter ones:
x 2y z = 4
8y 5z = 7
12 y 10 z = 8
Next we must eliminate y or z in either the second or third equation. It looks like we can
do this easily by adding 2 times the second equation to the third:
16 y 10 z = 14
12 y 10 z = 8
___________________
4y
= 6
The equivalent system of equations is now formed by replacing either the second or the
third equation with this new one that involves only y:
x 2y z = 4
8 y 5 z= 7
4y
= 6
The last equation quickly gives:
y = 1.5
We then substitute this value into the second equation to get:
8 ( 1.5 ) 5 z = 7
12 5 z = 7
5 z= 5
z= 1
Finally, we substitute z = 1 and y = 1.5 into the first equation to get:
x 2 ( 1.5 ) 1 = 4
x 3 1= 4
x 2= 4
x= 2
The solution to the system is then x = -2, y = 1.5, z = 1.
Once again, we should check this answer by substituting the three numbers into the
original system of equations to verify that they work. This is left to the reader to
complete.
For practice, let's do one more system of three equations with three variables:
2 x y 3 z = 14
4 x 2 y z = 12
6 x y 4 z = 22
Although we could easily eliminate y or z from two of the equations, let's eliminate x
from the second and third equations. Begin by multiplying the first equation by -2 and
adding the result to the second equation:
4 x 2 y 6 z = 28
4 x 2 y z = 12
_____________________
4 y 7 z = 40
Next, multiply the first equation by -3 and add the result to the third equation:
6 x 3 y 9 z = 42
6 x y 4 z = 22
_____________________
2 y 5 z = 20
The new, equivalent system of equations consists of these two equations where x has
been eliminated together with one of the original equations. Let's choose the first
equation so that the system becomes:
2 x y 3 z = 14
4 y 7 z = 40
2 y 5 z = 20
Now we try to eliminate y or z from the second or third equation. It looks to be easiest to
take -2 times the last equation and add it to the second equation:
4 y 10 z = 40
4 y 7 z = 40
__________________
3z = 0
Our equivalent system of equations is now the same as before except that we replace
either the second of third equation by 3z = 0.
2 x y 3 z = 14
4 y 7 z = 40
3 z= 0
Once we get this pattern where two x's have been eliminated and 1 y has been eliminated,
it is easy to find the solution. Start with the shortest equation and work back to the
longest. Thus we first solve 3z = 0 to get:
z=0
Then we substitute this value into the second equation to get:
4 y 7 ( 0 ) = 40
4 y = 40
y = 10
Finally, we substitute 10 for y and 0 for z in the first equation to get:
2 x 10 3 ( 0 ) = 14
2 x 10 = 14
2 x= 4
x= 2
Thus, the solution to our system of equations is x = -2, y = 10, and z = 0. It is suggested
that you substitute these values into the original equations to verify that they do indeed
provide a solution.