Math 10 Week 15: Solving Systems by Elimination

Elimination is a sometimes easier method of solving systems, depending on the two equations used. You would use this method if there is no variable without a coefficient.

To start, here’s an example: 5x + 3y = 10 and 5x – 2y = -15.

Since all the variables have coefficients, substitution is not the ideal method to use for this system.

Elimination is where you make a zero pair in one of the variables, and it has to be the x or y value and not the constant.

So for this equation, we can make a zero pair by multiplying one of the equations by -1, in this case it doesn’t matter.

I decided to multiply the second equation (5x – 2y = -15) with -1 so we can make everything positive besides the x value.

Now we have 5x + 3y = 10 and -5x + 2y = 15

Now you can add the two together, I suggest putting them on top of each other like this:

5x + 3y = 10

-5x +2y = 15


now add them together and you get 5y = 25, and all you do now is algebra, put y by itself by dividing both sides by 5, and you get y = 5

Now you put y = 5 into any of the equations, in my preference, the negative one, to get 5x – 2(5) = -15

5x – 10 = -15

now add the 10 on both sides to get 5x = -5, then divide both sides by 5 to get x = -1

And that’s how you use the method of Elimination, next time, we will be doing…. some sequences, see ya next time!

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