Another way of solving systems of linear equations is called the elimination or addition method. It is called the elimination method because after combining the two linear equations using addition, one of the two variables is canceled out. Thus, giving way to a simpler linear equation that is expressed in just one variable which can easily be solved.
Example 1: Solve the system of equations below using the elimination or addition method.
$$\eqalign{
2x + 5y &= 17 \cr
x - 5y &= 1 \cr} $$
Observe, the coefficients of the y-terms are opposite each other. They have the same absolute value but opposite in signs. If we add \(+5y\) and \(-5y\) they will cancel each other, that is, \(\left( { + 5y} \right) + \left( { - 5y} \right) = 0\).
With that in mind, let's add the two equations thereby eliminating the y-terms. What we have left is an equation in one variable which is in x.
$$\eqalign{
2x + 5y &= 17 \cr
x - 5y &= 1 \cr } $$
$$\rule{60pt}{1pt}$$
$$\eqalign{
3x &= 18 \cr
x &= 6\cr} $$
Now back-substitute the newly obtained value of x to either of the two given equations then solve for y. Let's select the second equation because it is simpler.
$$\eqalign{
x - 5y &= 1 \cr
6 - 5y &= 1 \cr
- 5y &= 1 - 6 \cr
- 5y &= - 5 \cr
y &= 1 \cr} $$
Verify that our solution is indeed \(\left( {1,6} \right)\).
Example 2: Solve the system of equations using the elimination method.
$$\eqalign{
3x - 6y &= 27 \cr
x + 2y &= 17 \cr} $$
This system of equations is not ready for addition yet. As you can see, adding the x-terms or y-terms won't eliminate either because their coefficients are not opposite each other. That means we have to perform a preliminary step first to force them to have opposite coefficients.
There are a few ways to accomplish this. However, the most obvious one is to multiply the bottom equation by 3 because y-terms are already opposite in sign. We just need to make their absolute value the same. Remember, multiplying the entire equation by a number does not change its meaning.
$${\color{red}{3}}\left( {x + 2y = 17} \right) \to 3x + 6y = 51$$
By making the coefficients of the y-terms opposite each other, \(-6\) and \(+6\), we can now proceed by adding the two equations: the original first equation and the transformed second equation.
$$\eqalign{
& 3x - 6y = 27 \cr
& 3x + 6y = 51 \cr} $$
$$\rule{60pt}{1pt}$$
$$\eqalign{
6x &= 78 \cr
x &= 13 \cr} $$
Select any of the equations to solve for y. For practice, let's choose the more complicated one which is the first equation.
$$\eqalign{
3x - 6y &= 27 \cr
3\left( {13} \right) - 6y &= 27 \cr
39 - 6y &= 27 \cr
- 6y &= 27 - 39 \cr
- 6y &= - 12 \cr
y &= 2 \cr} $$
Verify that the final answer should be \(\left( {13,2} \right)\).
Example 3: Solve the system of equations using the addition method.
$$\eqalign{
2x + 3y &= - 4 \cr
3x + 5y &= - 7 \cr} $$
This is an example where we have to multiply both equations by some numbers such that when we add them either the x-terms or y-terms are eliminated.
If we want to eliminate the x-terms, we will have to multiply the top equation by \(-3\) and the bottom equation by \(2\).
For the top equation:
$$ {\color{red}{- 3}}\left( {2x + 3y = - 4} \right) \to - 6x - 9y = 12$$
For the bottom equation:
$${\color{red}{2}}\left( {3x + 5y = - 7} \right) \to 6x + 10y = - 14$$
Notice the coefficients of the x-terms are additive inverses of each other. Let's add the transformed equations.
$$\eqalign{
- 6x - 9y &= 12 \cr
6x + 10y &= - 14 \cr} $$
$$\rule{60pt}{1pt}$$
$$\boxed{y = - 2}$$
To find the value of x, we pick the top equation and substitute \(y=-2\).
$$\eqalign{
2x + 3y &= - 4 \cr
2x + 3\left( { - 2} \right) &= - 4 \cr
2x - 6 &= - 4 \cr
2x &= - 4 + 6 \cr
2x &= 2 \cr} $$
$$\boxed{x = 1}$$
The final solution is \(\left( {1, - 2} \right)\).