Solving Equations with Multiple Variables Examples

Suppose a rectangle has a length of 5 cm and a width of 2 cm. What is the area of the rectangle?

We use the formula A = lw, substituting 5 for l and 2 for w, to find that

A = (5)(2) = 10 cm²

Sometimes these problems are a little more interesting. See how we said "interesting" instead of "complicated?" Man, are we good. Instead of getting the answer immediately, substituting in numbers may leave us with an equation in one variable that needs to be solved to find the answer.

Find the width of a rectangle with a length of 3 in and an area of 45 in².

Use the formula for width, substituting 3 for l and 45 for A.

Solve the formula C = 2πr for r.

Divide both sides of the equation by 2π and then call it a day:

If the circumference of a circle is 40π, what is the radius of the circle?

There are two ways to do this problem. Unfortunately, both of them involve dealing with that nasty little pi symbol.

Way 1: We know C = 2πr. We can substitute the value 40π for C to get the equation 40π = 2πr, which we can solve to find that r = 20.

Way 2: In the previous example, we solved the formula C = 2πr for r to get the new formula .

If we evaluate this new formula for C = 40π, we also find that r = 20.

If 5x – 9y = 12, what is x when y = 3?

Way 1: Substitute y = 3 into the equation to get:

5x – 9(3) = 12

And then solve to find that .

Way 2: Solve the original equation for x to get a formula that gives x in terms of y:

Then substitute 3 for y:

Either way, we do the same amount of work. The only difference is whether we rearrange things before or after we substitute in numbers. It's the same way you get your pants on whether you put your right leg or left leg into them first. Or maybe you're one of those people who leaps into the air and puts them on both legs at once. We've heard about you.