Algebra: Equations with Multiple Variables

Equations with Multiple Variables

So far, when I've asked you to solve an equation for a variable, it was pretty obvious which one I was talking about. For example, to solve the equation 3x + 2 = 23 , you'd solve for (isolate) the x variable. Why? Because it's the only variable in there! That x is enjoying all the attention, like the only girl in an all-boys school.

I need to add another skill to your equation-solving repertoire that will be extremely important in Graphing Linear Equations: how to solve for a variable when there's more than one variable in the equation. Don't worryyou don't need to learn a whole new set of steps to follow or anything; it's basically done the same way you're used to. Here's the only difference: When you're finished, you'll have variables on both sides of the equation, rather than just one side.

Example 4: Solve the equation -2(x - 1) + 4y = 5 for y.

Solution: Start by simplifying the left side of the equation.

  • -2x + 2 + 4y = 5

Now it's time to separate the variable term. Since you're trying to isolate the y, eliminate every term not containing a y from the left side of the equation. In other words, add 2x to, and subtract 2 from, both sides of the equation.

  • 4y = 2x + 3

It might have felt weird to move that -2x term, but it had to go, since you're solving for y, not x. All that's left to do now is eliminate y's coefficient by dividing both sides by 4.

  • y = 2x + 34
You've Got Problems

Problem 5: Solve the equation 9x + 3y = 5 for y.

Even though that answer is correct, it will make more sense in Graphing Linear Equations if you rewrite it slightly. Any time two or more things are added or subtracted in the numerator of a fraction, you can break that fraction into smaller fractions, each of which will contain one term of the original numerator and a copy of the denominator. (If the terms contain variables, just stick them to the right of the fraction.)

  • y = 24x + 34
  • y = 12x + 34
CIG Algebra


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