So far, we’ve solved a system of equations by graphing and by substitution. Today, we tackle the challenging method of solving a system of equations by elimination.
You’ve probably already figured out that something is going to get eliminated, and you’re right. When we solve by elimination, we eliminate one variable in order to solve for the other.
Let’s consider the following system of equations.
x + 3y = 6
2x – 3y = 3
The first thing we’re going to do is add these two lines together.
x + 2x = 3x
3y + (-3y) = 0
6 + 3 = 9
This leaves us with the new equation: 3x = 9. Notice that the y terms cancelled each other out. This is the elimination this method is named for.
When we solve 3x = 9, we find that x = 3. Now we plug 3 in for x in either equation. I’m going to use the first equation for this example.
3 + 3y = 6
Now I solve for y as usual.
3y = 3
y = 1
The solution, the one point that lies on both lines, is (3, 1).
Generally, when you look at a system of equations, one of the methods will jump out at you as the easiest method to use for the problems. For systems where you have an x or a y with no coefficient (the number in front of the variable), you’re probably best off using substitution. If you have the positive of a term, and its exact negative in the other equation, go for elimination.
