Solving algebra equations is difficult for many. If you struggle with algebra solving equations can be improved using this lesson and practice.

Equations with more than one variable

Sometimes you might see an equation with more than one variable, like this one:

2x + 6y -10 = 38

If an expression has more than one variable, you won't be able to simplify it all the way—there's not enough information. Instead, problems with equations that have multiple variables will usually ask you to solve for one of the variables. You'll simplify it as much as you can, with the variable you're solving for on one side of the equation and any other numbers and variables on the other. Let's simplify the expression above: 2x +6y - 10 = 38.

We can't do anything with the order of operations, so let's start cancelling things out. We want x alone on the left side, so we'll try to get everything else on the right.

2x + 6y - 10 = 38

First, we'll cancel out -10. The opposite of -10 is 10, so we'll add 10 to both sides.-10 + 10 is 0, and 38 + 10 is 48.

2x

+ 6y

- 10

=

38

+ 10

+ 10

Next, let's get rid of 6y. We'll subtract it from both sides. 6y - 6y is 0. Because there's nothing to subtract it from on the other side, we'll just write -6y on the right. (Confused? It's like we subtracted 6y from nothing, or 0—and 0 - 6y is -6y.)

2x

+ 6y

=

48

- 6y

- 6y

Now we have to get rid of the 2 in 2x. Because 2x is another way of saying 2 ⋅ x, we'll divide both sides by 2 to get x alone on the left. 2x / 2 is x, and (48 - 6y) / 2 is 24- 3y.

2x

=

48

- 6y

2

2

That's all it takes! The expression isn't fully simplified—we still don't know the numerical value of x and y—but it's simplified enough because we can say that x equals 24 - 3y.

x = 24 - 3y

Remember, your goal with problems like this isn't to completely simplify the expression—it's to find the value of one of the variables.

It is actually possible to solve for two variables when you have more than one equation with the same variables. This is called a system of equations. We actually use systems of equations in our lesson on distance word problems, but we don't discuss how they work in general. To learn more about systems of equations, check out this video from Khan Academy.

Watch the video below to see this example problem solved.

Practice!

Problem 1

Solve for r.

88q + 4r - 3 = 5

Problem 2

Solve for s. (Hint: your final answer will be a fraction with a denominator of r.)