# Elimination Method (Systems of Linear Equations)

The main concept behind the **elimination method** is to create terms with **opposite coefficients** because they cancel each other when added. In the end, we should deal with a simple linear equation to solve, like a one-step equation in [latex]x[/latex] or in [latex]y[/latex].

## Two Ideal Cases of the Elimination Method

I can summarize the “big” ideas about the elimination method when solving systems of linear equations using the illustrations below. Here I present two ideal cases that I want to achieve during the solving process. Take a look at them and hopefully, it makes sense. Otherwise, go directly to the six (6) worked examples to see how actual problems are being solved.

**Case 1**: By Adding the Two Equations, the Variable “[latex]x[/latex]” is Eliminated

The coefficients of **variable** [latex]x[/latex] are opposites.

**Case 2**: By Adding the Two Equations, the Variable “[latex]y[/latex]” is Eliminated

The coefficients of **variable** [latex]y[/latex] are opposites.

### Examples of Solving Systems of Linear Equations by Elimination Method

**Example 1:** Solve the system of linear equations by elimination method.

I have observed that adding the [latex]x[/latex]-column will not eliminate the variable [latex]x[/latex]. However, if I add the [latex]y[/latex]-column the variable [latex]y[/latex] disappears. This happens because the coefficients of [latex]y[/latex] are opposite of each other in terms of signs. Now, I will proceed with the second option.

After doing so, I end up with an easy equation.

I divide both sides by the coefficient of [latex]x[/latex] which gives the answer of [latex]x = 4[/latex].

The next step is to find the corresponding value of [latex]y[/latex]. This is easy to find since I already know what [latex]x[/latex] is. I will pick any of the two original equations, which in this case, I chose the top equation. Then, I will plug in the value of [latex]x = 4[/latex] to get [latex]y[/latex]. The process or procedure of solving for [latex]y[/latex] should be similar below.

Here I get [latex]y = – \,4[/latex]. The final answer in point notation is shown below.

Graphically, the solution looks like this.

**Example 2: **Solve the system of linear equations by elimination method.

This is quite interesting because no variables will cancel when added. What I want is to introduce a multiplier to one of the equations, or both, and then observe if I arrive at some coefficients that only differ in signs.

There are a

Multiplying the entire equation by any nonzero number **does not change** its original meaning. What will change is just its form. I call this process, equation “revision” or “modification”.

This is a one-step equation so I solve [latex]y[/latex] by dividing both sides by its coefficient.

Great! I obtained the value [latex]y = 2[/latex]. Next, I will solve [latex]x[/latex] using back substitution using either of the original equations. For this, I will utilize the top equation because it is less complicated.

I obtained the value [latex]x = – 1[/latex]. I can now write the final answer as the ordered pair written below.

The graph below verifies that our solution is correct.

**Example 3:** Use the method of elimination or linear combination to solve.

There’s some twist on this problem because the coefficients of [latex]x[/latex] variables are exactly the same, **both** [latex]- 2[/latex]. The only thing I need to fix here is to make one of them positive. Now, I decided to multiply the top equation by [latex]âˆ’ 1[/latex]. It should also work just fine if I multiply the bottom by [latex]âˆ’ 1[/latex].

You should see that the plan works since adding the [latex]x[/latex]-column results to the cancellation of [latex]x[/latex].

I solved the value of [latex]y[/latex] by dividing both sides by [latex]âˆ’ 17[/latex] which results to [latex]y = 3[/latex]. This time, I will back solve the value of [latex]x[/latex] using the bottom equation because I know what [latex]y[/latex] is.

After a few steps in solving the equation above, I arrive at [latex]x = 2[/latex]. The final answer as an ordered pair is shown below.

Indeed, the two lines intersect at the point that we found in our calculations.

**Example 4:** Use the method of elimination or linear combination to solve.

This example follows along the line of example 3 where we have exactly the same coefficients. I see that variable [latex]y[/latex] both have coefficients of [latex]8[/latex]. So, I will need to tweak it a bit in order to make their signs opposite. I now have two options on how to proceed. I can multiply the top equation by [latex]âˆ’ 1[/latex] or the bottom by [latex]âˆ’ 1[/latex] as well. For this exercise, I choose to do the latter.

Applying the [latex]âˆ’ 1[/latex] multiplier on the bottom equation and adding them together results to [latex]y[/latex] going away.

Solve the simple equation that arises from it.

I got [latex]x = 4[/latex] by dividing both sides by [latex]âˆ’ 9[/latex]. The next obvious step is to solve for the other variable [latex]y[/latex] using back substitution. Pick any of the original equations, plug [latex]x = 4[/latex], and you will get [latex]y[/latex] in no time.

The answer is [latex]y = – \,1[/latex]. The final answer in the ordered pair form is shown below.

The graphical solution looks like this.

**Example 5: **Use the method of elimination or linear combination to solve.

This type of problem requires us to simultaneously multiply both the top and bottom equations by some number in order to generate coefficients with opposite signs.

If I decide to eliminate [latex]x[/latex], I can multiply the top equation by [latex]âˆ’ 2[/latex] and the bottom by [latex]9[/latex]. By doing so, I should end up with [latex]x[/latex] terms, [latex]18x[/latex] and [latex] – 18x[/latex], respectively, which would cancel when added together.

For this exercise, I want to eliminate [latex]y[/latex]. Therefore, I will multiply the top by [latex]5[/latex] and the bottom by [latex]3[/latex].

As predicted, I was able to get rid of [latex]y[/latex] which leaves us with a simple equation to deal with.

You should arrive at [latex]x = 1[/latex]. Proceed on solving the other variable which is [latex]y[/latex]. Plug in [latex]x = 1[/latex] then solve for [latex]y[/latex].

I got [latex]y = – \,2[/latex]. Putting it together, our final answer is the ordered pair below**.**

The graphical representation of the two lines intersecting at the solved point is…

**Example 6: **Use the method of elimination or linear combination to solve.

This last example is very similar to the previous one. As it stands, no variables will be eliminated after adding the columns of [latex]x[/latex] and [latex]y[/latex]. However, I can eliminate the [latex]x[/latex] variables by multiplying the first equation by [latex]5[/latex] and the second by [latex]âˆ’ 4[/latex], and then add them together. The rest is history!

I will end up solving a simple equation as shown.

I now have [latex]y = 5[/latex] after dividing both sides by [latex]8[/latex]. I will then substitute this value of [latex]y[/latex] to any of the original equations to solve for the corresponding [latex]x[/latex]-value.

This yields an answer of [latex]x = – \,6[/latex]. The final answer should be [latex]\left( {x,y} \right) = \left( { – \,6,5} \right)[/latex] .

This point is where the two lines intersect, as shown below.

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