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# Equation with two rational expressions (old example)

Video transcript

Solve the equation and
find excluded values. And what they're talking about,
about finding excluded values, is we need to think about what
values would make these either side of this equation undefined. And the reason why
we want to do that is because as we
manipulate this, we might lose things
in the denominator. And then we might
get some answer. But if it's one of
those things that made the original, the original
expressions, or either side of the original
equation undefined then that wouldn't be
a legitimate solution. So that's what they're talking
about, the excluded values. So what values do we have
to exclude here right from the get go? Well 4 over p minus
1 won't be defined if p is 1 because if p is
1, then this, then you're going to be dividing by 0. And that's undefined. So we know that p
cannot be equal to 1. And over here, if p
was at negative 3, then this denominator would be
0 and it would be undefined. And so p cannot be equal
to 1 or negative 3. So these right here are
our excluded values. So now let's try to solve. Let's try to solve
this equation. And I'm going to
rewrite it over here. So we have 4 over p minus 1
is equal to 5 over p plus 3. So the first thing we
could do, especially because we can assume now that
neither of these expressions are 0. And this is going to
be defined, since we've excluded these values of
p, is to get the p minus 1 out of the denominator. We can multiply the left
hand side by p minus 1. But remember, this
is an equation. If you want them to
continue to be equal, anything you do
left hand side, you have to do to the
right hand side. So I'm multiplying by p minus 1. Now I also want to get this p
plus 3 out of the denominator here on the right hand side. So the best way to do that is
multiply the right hand side by p plus 3. But if I do that to
the right hand side, I also have to do that
to the left hand side. p plus 3. And so what happens? We have a p minus 1 in
the numerator, p minus 1 at the denominator. These cancel out. So you have just a 1
of the denominator, or you have no
denominator anymore. And the left hand
side simplifies to 4 times p plus
3, or if you were to distribute the
4, 4 times p plus 3. So that is 4p plus 12. And then the right
hand side, you have plus 3 canceling
with a p plus 3. This is p plus 3
divided by p plus 3. And all you're left with
is 5 times p minus 1. If you distribute the
5 you get 5p minus 5. And now this is a pretty
straightforward linear equation to solve. We just want to isolate
the p's on one side and the constants on the other. So let's subtract
5p from both sides. I'll switch colors. So let's subtract
5p from both sides. And we get on the
left hand side, 4p minus 5p is
negative p plus 12. Is equal to, these cancel
out, is equal to negative 5. And then we could subtract
12 from both sides. And we get, these cancel out,
we get negative p is equal to, negative 5 minus
12 is negative 17. And we're almost done. We can multiply both
sides by negative 1 or divide both
sides by negative 1 depending on how
you want to view it. And we get, negative
one times negative p is. So let me just scroll
down a little bit so I have a little
bit of real estate. That's positive
p is equal to 17. p is equal to 17. And let's verify that
this really works. Well, it wasn't one of our
excluded values, but just in case, let's verify
that it really works. If we go, if we have p is 17. We get 4 over 17 minus 1,
needs to be equal to 5 over 17 plus 3. I'm just putting 17 in for p,
because that's our solution. So this is the same
thing as 4 over 16, needs to be the same
thing as 5 over 20. Or 4/16 is the
same thing as 1/4, and that needs to be
the same thing as 5/20, which is the same thing as 1/4. So it all checks out. So these are excluded values. And lucky for us, this
wasn't one of them.