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 } $$

\(\hrule{60pt}{1pt}\)

$$\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 \(\left( {1, - 2} \right)\).