# Linear Combinations

## Simple Addition

The process for solving a "**system of linear equations**", i.e. two or more lines, is similar to the process for solving a single equation with one variable. In fact, it's nothing more than another application of the Additive Property of Equality. Here's a simple example:

*x* + *y* = 4*x* - *y* = 6

I wrote the two equations with one over the other because it helps you to see how they're going to be combined. If I add together the right hand sides and the left hand sides I get

*x* + *y* = 4*x* - *y* = 6

2*x* + (*y* - *y*) = 4 + 6

2*x* = 10*x* = 5

Now that we have the *x* value of the solution, we can get the *y* value by substituting 5 into either of the original equations. If we use the first one we get

5 + *y* = 4*y* = 4 - 5 = -1

So our answer is (5, -1). You can check this by substituting *x* = 5 and *y* = -1 into both of the original equations and confirming that you get a true statement back.

## Multiplication

Most of the time you won't have a situation where one of the variables is the negative of the other so that they automatically cancel each other out. In this case, you'll have to multiply one or both of the equations by some number first. For example,

3*x* + 2*y* = 0*x* - 3*y* = -11

I'm going to try to cancel out the *y*'s although you could just as easily work with the *x*'s. I see a 2 and a 3 as the *y* variables's coefficients so I'm going to try to make them both equal to 6. (Why 6? Because it's the least common multiple of 2 and 3. If the numbers are large or you just don't want to bother finding the least common multiple you can always just use the product of the numbers.) To do this I have to multiply the top equation by 3 and the bottom one by 2.

3 · (3*x* + 2*y* = 0)

2 · (*x* - 3*y* = -11)

This gives us two equations that we can add together and have things cancel.

9*x* + 6*y* = 0

2*x* - 6*y* = -22

11*x* = -22*x* = -2

Now the problem goes just like the prevoius one. Substitute *x* = -2 into one of the original equations, say the first one, and we get

3 · -2 + 2*y* = 0

-6 + 2*y* = 0

2*y* = 6*y* = 3

So our solution is the point (-2, 3).