# 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.