# Question Video: Finding the Solution to a Rational Equation Containing Three Algebraic Fractions

Solve (3/(πΒ² β 4)) + (1/(π + 2)) = 2/(π β 2).

02:21

### Video Transcript

Solve three over π squared minus four plus one over π plus two equals two over π minus two.

We have three algebraic fractions. And there are a number of ways we can solve this. The key to any of these methods, though, is spotting that π squared minus four could be factored. We can write it using the difference of two squares. We write it as π minus two times π plus two. And then we notice that this is the product of our other two denominators. So we could create a common denominator of π minus two times π plus two and gather all our terms on the left-hand side. Alternatively, we could subtract one over π plus two from both sides of our equation. Letβs see what that would look like.

Our equation becomes three over π squared minus four equals two over π minus two minus one over π plus two. Next, weβll multiply the numerator and denominator of our first fraction by π plus two and of our second by π minus two, creating a common denominator of π minus two times π plus two. The numerators on the right-hand side become, respectively, two times π plus two and one times π minus two. And in fact, we saw we could write the denominator on the left-hand side as π minus two times π plus two.

Now, the denominators are equal on the right-hand side. Weβre going to subtract one times π minus two from two times π plus two. And the expression on the right-hand side becomes two times π plus two minus one times π minus two over π minus two times π plus two. Notice now that the denominators of these two fractions are equal. And weβre told the fractions themselves are equal. So this means our numerators must be equal also. So three must be equal to two times π plus two minus one times π minus two.

Letβs distribute these parentheses, remembering that, on our second set of parentheses, weβre multiplying everything by negative one and our equation becomes three equals two π plus four minus π plus two. We simplify the right-hand side to get π plus six. And we now see we have quite a simple equation that we can solve for π. We subtract six from both sides, giving us π equals negative three. So the solution to our equation is π equals negative three.

Remember, we could check the solution by substituting it back into the original equation and making sure that both sides are then equal.