Equations with rational expressions (example 2) | Mathematics III | High School Math | Khan Academy

So we have a nice little equation here that has some rational expressions in it. And like always, pause the video and see if you can figure out which X's satisfy this equation.

Alright, let's work through it together. Now, when I see things in the denominator like this, my instinct is to try to not have denominators like this. And so what we could do is to get rid of this x - 1 in the denominator on the left-hand side. We can multiply both sides of the equation by (x - 1).

So we're going to multiply both sides by (x - 1). Once again, the whole point of doing that is so that we get rid of this x - 1 in the denominator right over here. Then, to get rid of this x + 1 in the denominator over here, we can multiply both sides of the equation by (x + 1).

So, (x + 1) multiply both sides by (x + 1). And so what is that going to give us? Well, on the left-hand side, that is going to be (x - 1)/(x - 1), which is just going to be 1 for the x's where that’s defined, for x not being equal to 1.

So we're going to have (x + 1) * (-2x + 4). Let me write that down. So we have x + 1 times (-2x + 4) is going to be equal to now. If we multiply both of these times 3/(x + 1), the x + 1 is going to cancel with the x + 1, and we're going to be left with 3 * (x - 1).

So that is going to be 3x - 3. Then, minus (-1) times both of these, so -1 * (x - 1) * (x + 1). All I did is I multiplied, took (x - 1) * (x + 1), and multiplied it times each of these terms. When I multiplied it times this first term, the x + 1 and the x + 1 canceled, so I just had to multiply 3 * (x - 1).

For the second term, I just multiply it times both of these. And now you might recognize this. If you have something like (x + 1) * (x - 1), that's going to be (x^2 - 1). So I could rewrite all of this right over here as being equal to (x^2 - 1).

And once again, that's because this is the same thing as (x^2 - 1). Since I'm subtracting an (x^2 - 1), actually let me just—I don't want to do too much in one step, so let's go to the next step.

I could multiply this out. So I could multiply x * (-2x), which would give us -2x^2. x * 4, which is going to give us +4x. I can multiply 1 * (-2x), so I'm going to subtract 2x. Then, 1 * 4, which is going to be +4.

That is going to be equal to 3x - 3, and then we can distribute this negative sign. So we can say - (x^2 - 1). Over here, we can simplify it a little bit. This is going to be— that is 4x - 2x, so that would be 2x.

So this is simplified to—let's see. Well, this is—we have a -3 and a +1. So those two together are going to be equal to -2. We can rewrite everything as—do a neutral color now—-2x^2 + 2x + 4 is equal to (x^2 + 3x - 2).

Now we can try to get all of this business onto the right-hand side, so let's subtract it from both sides. We'll add x^2 to both sides; that gets rid of this negative x^2. Then we subtract 3x from both sides and add 2 to both sides.

We will be left with—let's see—(-2x^2 + x^2) which is -x^2. Then, (2x - 3x) is -x, and then (4 + 2) is 6, which is going to be equal to 0. I don't like having this negative on the x's, so let's multiply both sides by -1.

If I do that, if I just take the negative of both sides, I'm going to get—I'm going to get positive x^2 + x - 6 is equal to 0. And we're making some good progress here, so we can factor this.

Actually, let me just do it right over here so that we can see the original problem. If I factor this, what two numbers (their product is -6) and they add up to the coefficient of 1 on the first-degree term? Well, positive 3 and negative 2 work.

So I can rewrite this as (x + 3) * (x - 2) is equal to 0. Did I do that right? Yeah, 3 * -2 is -6, and 3x - 2x is +x. Alright, so I just factored. I just wrote this quadratic in factored form.

The way that you get this equaling zero is if either one of those equals zero: (x + 3) = 0 or (x - 2) = 0. Well, this is going to happen if you subtract 3 from both sides; you get—that's going to happen if x is equal to -3. Or over here, if you add 2 to both sides, x is equal to 2.

So either one of these will satisfy, but we want to be careful. We want to make sure that our original equation isn't going to be undefined for either one of these. -3 does not make any of the denominators equal to zero, so that's cool.

And positive 2 does not make any of the denominators equal to zero, so it looks like we're in good shape. There are two solutions to that equation. If one of them made any of the denominators equal to zero, then they would have been extraneous solutions.

They would have been solutions for some of our intermediate steps, but not for the actual original equations with the expressions as they were written. But this, we can feel good about because neither of these make any of these denominators equal to zero.