Main content

## Algebra II (2018 edition)

### Unit 6: Lesson 5

Solving rational equations# Rational equations intro

CCSS.Math:

When we have an equation where the variable is in the denominator of a quotient, that's a rational equation. We can solve it by multiplying both sides by the denominator, but we have to look out for extraneous solutions in the process. Created by Sal Khan.

## Video transcript

- [Instructor] Let's say we wanna solve the following equation for x. We have x plus one over nine
minus x is equal to 2/3. Pause this video and
see if you can try this before we work through it together. All right now let's work
through this together. Now, the first thing
that we might wanna do, there's several ways that
you could approach this, but the thing I like to do is get rid of this x here in the denominator. And the easiest way I
can think of doing that, is by multiplying both sides of this equation by nine minus x. Now, when you do that, it's important that you
then put the qualifier that the x cannot be equal to the value that would have made this
denominator zero 'cause clearly if somehow you do all this
algebraic manipulation and you got x is equal to nine that still wouldn't be a valid solution 'cause if you were to substitute nine back into the original equation you'd be dividing by
zero in the denominator. So, let's just put that right over here, x cannot be equal to nine. And so then, we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not
equal nine, if we multiply, and divided by nine
minus x they cancel out, and we'll just be left with an x plus one, and on the right-hand side, if you multiply 2/3 times nine minus x, we get 2/3 times nine is six and then 2/3 times negative
x is negative 2/3 x and once again, let's remind ourselves, that x cannot be equal to nine. And then we can get all of
our x's on the same side so let's out that on the left. So let's add 2/3 x to both sides. So plus 2/3, 2/3 x plus 2/3 x, and then, what do we have? Well, on the left-hand side we have one x which is the same thing
as 3/3 x plus 2/3 x is going to give us 5/3 x
plus one is equal to six, and then these characters cancel out. And then we can just
subtract one from both sides, and we get 5/3 x, 5/3 x is equal to five. And then, last but not least, we can multiply both
sides of this equation, times the reciprocal of 5/3 which is of course 3/5, and I'm doing that so I just have an x isolated
on the left-hand side. So times 3/5, and we are left with 3/5 times 5/3 is of course equal to one. So we're left with x is equal
to five times 3/5 is three. And so we're feeling pretty
good about x equals three, but we have to make sure
that that's consistent with our original expression. Well if we look up here, or if you substitute back x equals three, you don't get a zero in the denominator, x is not equal to nine. X equals three is consistent with that. So we should feel good about our solution. If we did all this algebraic manipulation and we get an x is equal to nine, then that still wouldn't
be a valid solution because it would have made the original expression
on the left be undefined.