### 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. Letβs 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, weβll 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. Weβre 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 weβre 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.

Letβs distribute these parentheses,
remembering that, on our second set of parentheses, weβre 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.