Equations with rational expressions | Mathematics III | High School Math | Khan Academy

So we have a nice little equation here dealing with rational expressions, and I encourage you to pause the video and see if you can figure out what values of x satisfy this equation.

All right, let's work through this together. The first thing I'd like to do is just see if I can simplify this at all, and maybe by finding some common factors between numerators and denominators or common factors on either side of the equal sign.

So let's factor all of these. All of the numerators and denominators, all the ones on the right-hand side are already done. So this thing up here I could rewrite as, let's see, what product is 21? What two numbers, when I take their product, is 21? Positive 21. So, they're going to have the same sign, and when I add them, I get -10.

Well, 7 and -3. So this could be written as (x - 7)(x - 3). This over here, both are divisible by three; I could rewrite this as 3(x - 4), and these are already factored.

So, the one thing that jumps out to me is I have (x - 4) in the denominator on the left-hand side and on the right-hand side. So if I were to multiply both sides by (x + 4)...actually, let me just—let me formally replace this with that.

And up here, it's not so obvious that it's going to be valuable for me to keep this factored form, so I'm just going to keep it in this yellow form, in the expanded out form. So let me just scratch that out for now because once I—well let me multiply by (x - 4).

So if we multiply both sides by (x - 4)—and once again, why am I doing this? It's so I get rid of the (x - 4) in the denominator. (x - 4) and then (x - 4) that and that cancels. That cancels.

And then we are left with, in the numerator, we are left with our (x^2 - 10x + 21) divided by 3 is equal to (x - 5).

Let's see now what we could do, and actually, I could have done it in the last step. I could multiply both sides by three. Multiply both sides—do that in another color just so it sticks out a little bit more.

So I can multiply both sides by three. So, multiply both sides by three on the left-hand side. That and that cancels, and I'll just be left with (x^2 - 10x + 21).

And I don't have a denominator anymore; my denominator is one, so I don't need to write it. It is going to be equal to (3(x - 5)). Let's distribute the three: (3 * x = 3x) and (3 * -5 = -15).

Now I can get this in standard quadratic form by getting all of these terms under the left side. The best way to do that, let's subtract (3x) from the right, but I can't just do it from the right; otherwise, the equality won't hold.

I have to do it from both sides if I want the equality to hold, and I want to get rid of this -5, so I can add 15 to both sides. So let's do that, and what we are left with—scroll down a little bit so we have a little more space.

What we are going to be left with is (x^2 - 13x + 36 = 0). All right, now let's see. We have this quadratic in standard form. How can we solve this?

So first thing, can we factor this? The product of two numbers 36, if I add them, I get -13. They're both going to be negative since they have to have the same sign to get their product to be positive.

And let's see, -9 and -4 seem to do the trick. So, ((x - 9)(x - 4) = 0). Well, that's going to happen if either (x - 9 = 0) or (x - 4 = 0).

Well, add 9 to both sides of this; this happens when (x = 9). Add 4 to both sides of this; this happens when (x = 4). So we could say that the solutions are (x = 4) or (x = 9).

So (x) is equal to 4 or (x) is equal to 9. But we need to be careful because we have to remember in our original expression (x - 4) was a factor of both denominators.

And so if we actually tried to test (x = 4) in the original equation—not one of these intermediary steps—in the original equation, I would end up dividing by zero right over here, and actually, to end up dividing by zero right over there as well.

So the original equations, if I tried to substitute 4, they don't make sense. So this is actually an extraneous solution; it's not going to be a solution to the original equation. The only solution is (x = 9).