Solving for a variable is an efficient way to find out the values that meet the constraints in a situation. Here is an example.

An elevator has a capacity of 3,000 pounds and is being loaded with boxes of two sizes—small and large. A small box weighs 60 pounds, and a large box weighs 150 pounds.

Let  be the number of small boxes and  be the number of large boxes. To represent the combination of small and large boxes that fill the elevator to capacity, we can write:

 

If there are 10 large boxes already, how many small boxes can we load onto the elevator so that it fills it to capacity? What if there are 16 large boxes?

In each case, we can substitute 10 or 16 for and perform acceptable moves to solve the equation. Or, we can first solve for :

This equation allows us to easily find the number of small boxes, , that can be loaded, by substituting any number of large boxes for .

Now suppose we first load the elevator with small boxes, say, 30 or 42, and want to know how many large boxes can be added for the elevator to reach its capacity.

We can substitute 30 or 42 for in the original equation and solve it. Or, we can first solve for :

Now, for any value of , we can quickly find by evaluating the expression on the right side of the equal sign.

Solving for a variable—before substituting any known values—can make it easier to test different values of one variable and see how they affect the other variable. It can save us the trouble of doing the same calculation over and over.