### Video Transcript

Consider the equation one over π₯
minus three minus two over π₯ plus three is equal to three plus π₯ over nine minus
π₯ squared. If you multiply both sides by π₯
squared minus nine and rearrange so that the right-hand side is zero, what equation
do you get? What can you say about the
solutions of the original equation?

Weβll start with what we were
given. And then weβre trying to multiply
both sides of this equation by π₯ squared minus nine. For the left-hand side, that means
weβll need to distribute π₯ squared minus nine to both of these terms. We multiply the numerator by π₯
squared minus nine. And our first term becomes π₯
squared minus nine over π₯ minus three. The second term looks like this,
two times π₯ squared minus nine over π₯ plus three. On the right-hand side, weβll have
three plus π₯ times π₯ squared minus nine over nine minus π₯ squared.

But we want to move the term on the
right-hand side to the left-hand side. If we subtract three plus π₯ times
π₯ squared minus nine over nine minus π₯ squared from both sides of the equation, we
get this. Weβve accomplished one of our
goals. And thatβs to get the right-hand
side to be equal to zero. But now we need to simplify these
three terms. To do this, letβs break them apart
and simplify each term by itself and then plug it back in.

Our first term is π₯ squared minus
nine over π₯ minus three. Here we have to notice a
pattern. π₯ squared minus nine is the
difference of two squares. And we rewrite the difference of
two squares as π plus π times π minus π. In our case, π₯ squared minus nine
is the same thing as saying π₯ squared minus three squared. And π₯ squared minus three squared
is the same thing as π₯ plus three times π₯ minus three. When we rewrite it this way, we see
that the π₯ minus three in the numerator and the denominator cancel out and that our
first term is only π₯ plus three. Weβll substitute that back in for
our first term.

Now when it comes to our second
term, notice that weβre doing subtraction. We canβt leave out that negative
sign. To accommodate for that, weβll say
negative two times π₯ squared minus nine over π₯ plus three. Again, we have the same difference
of squares. And we can rewrite the numerator as
negative two times π₯ plus three times π₯ minus three. π₯ plus three in the numerator and
the denominator cancel out. And weβre left with negative two
times π₯ minus three. We can distribute that negative
two. Negative two times π₯ equals
negative two π₯. Negative two times negative three
equals positive six. In place of our second term, weβll
write negative two π₯ plus six.

Now we have our third term, which
again has a negative in front of it. So weβll write negative one times
three plus π₯ times π₯ squared minus nine all over nine minus π₯ squared. Notice that we have π₯ squared
minus nine in the numerator and nine minus π₯ squared in the denominator. We can think of this as having a
positive π₯ squared minus nine and a positive nine minus π₯ squared. But if we multiply the numerator by
negative one, we would end up with negative π₯ squared and positive nine. So thatβs what we wanna do. We wanna multiply negative one by
π₯ squared minus nine. When we do that, we get negative π₯
squared plus nine.

Remember that when it comes to
multiplying, we can multiply the numerator in any order. So weβve just multiplied negative
one by π₯ squared minus nine. From there, we wanna switch these
two addends. Instead of saying negative π₯
squared plus nine, weβll say nine minus π₯ squared times three plus π₯ all over nine
minus π₯ squared. The nine minus π₯ squared in the
numerator and the nine minus π₯ squared in the denominator cancel out. And our third term is just three
plus π₯. In place of our third term, we
write three plus π₯. And we know that this is equal to
zero.

We now have π₯ plus three minus two
π₯ plus six plus three plus π₯. So weβll combine the like terms, π₯
minus two π₯ plus π₯. Well, if we add the two π₯s
together, weβll get positive two π₯ minus two π₯. So they cancel completely. Then, we add three plus six plus
three, which equals 12. Our simplified statement is that 12
equals zero. The first part of our question is
then 12 equals zero. The second half wants to know what
we can say about the solutions of the original equation. When weβre solving an equation and
we end up with a simplified form thatβs an untrue statement β 12 equals zero is
untrue β we can say that there are no solutions to the equation. There is no value for π₯ that will
make that statement true because zero is never equal to 12.