## Solving Linear Systems by Substitution

Because we’re working with variables, we often have to substitute values in for these variables. You’re already used to this. For example, if we were trying to solve 2x – 1 = 11, and you had come up with an answer of x = 6, you could check your answer by substituting in 6 for x in the equation 2x – 1 = 11. We are able to do that because we know x = 6.

So suppose we’re given the equations y = -3x + 1 and 4x – 5y = 33. If we put both of these equations into slope-intercept form, we’ll see that the slopes are different, so there will be only one solution. Now we’re going to solve for this solution using substitution.

We can substitute -3x + 1 in for y if we’d like, since we know they are equal. This gives us the following:

4x – 5y = 33
4x – 5(-3x + 1) = 33

From this point, we can solve for x normally like we have many times before.

4x + 15x – 5 = 33
19x – 5 = 33
19x = 38
x = 2

Now we can substitute 2 in for x in either of our original equations. Let’s use y = -3x + 1 for example:

y = -3x + 1
y = -3(2) + 1
y = -6 + 1
y = -5

Now we have x = 2 and y = -5, for an answer of (2, -5). As always, we should check our answer by putting these values into both of the original equations to make sure everything checks out. Here we go:

y = -3x + 1
-5 = -3(2) + 1
-5 = -6 + 1
-5 = -5

4x – 5y = 33
4(2) – 5(-5) = 33
8 + 25 = 33
33 = 33

Everything comes out right, so we know we have the right answer.