 So one of the goals of algebra is to solve every possible type of equation and rational expressions give rise to rational equations. So again, quick recap, a rational expression is essentially a fraction involving a variable quantity. So for example, 3 over x, or x plus 8 over 4, or something horrible like that, and some more horrible complicated expressions as well. And so a rational equation is going to be an equation that includes a rational expression. So how do we solve rational equations? Well, because we're looking at rational expressions, it is absolutely important to begin by identifying the forbidden solutions. Anything that makes a denominator equal to 0, we cannot allow as a solution. Other than that, probably the best way to proceed is nobody likes fractions. So what we can do is we can multiply through by all of the denominators. And that's going to eliminate all of our fractions. And then we'll have an equation of some form, which we'll hopefully be able to solve. So for example, let's take a look. Solve 1 over x minus 1 over x plus 4 equals 1 over 6. And so the first thing to do is you want to identify the forbidden values, things that will make a denominator equal to 0. So here I see I have denominator x. So I know that x can't be allowed to be 0. I have denominator x plus 4. So that says x can't be allowed to be negative 4. Denominator 6 is never going to be equal to 0. So I don't have to worry about that. So now our denominators are x, x plus 4, and 6. So if I multiply every term by their product, 6x, x plus 4, I'll eliminate all of the denominators. So I'll do that. Here's my original equation. I'll multiply everything by 6x, x plus 4. And I can cancel factors out. This denominator x will cancel with that. x plus 4, 6 is cancel. And I'll end up with 6 times x plus 4, 6x times 1, 1 times x times x plus 4. And I'll add it with a new equation. And I can expand and multiply things out. And I have a quadratic equation. So I want to get all terms over onto the right-hand side. And I can waste time trying to fact. And now I don't want to do that. I want to use the quadratic formula here. And that'll give me my solutions. x equals a is 1, b is 4, c is negative 24. I'll drop those into the quadratic formula. And I'll let the arithmetic dots clear. And I'll end up with my two solutions. Now remember, I'm not ever going to be allowed to use 0 or negative 4 as a solution. So I have to worry that one of our solutions, one or both solutions, could be my forbidden values. But squared of 112 is not a whole number. It's an irrational number. So neither of these two are going to be our forbidden values. So both of these solutions are valid. Well, let's take a look at a different one. So here's a problem. 3 over x plus 2, 3 over x minus 2 equals x over x squared minus 4. And again, the first thing I want to do is find the forbidden values, things that are going to make the denominator 0. So here, x plus 2, well, what's going to make x plus 2 equal to 0? Well, x equals negative 2. So I don't allow that. My next denominator, x minus 2 equals 0. And again, what's going to make this equal to 0? x equals 2 will make that equal to 0. So I don't allow that. x squared minus 4 equal to 0. And so I'll try and solve that denominator equal to 0. And so I can get x squared equals 4. x equals plus or minus the square root of 4. And that's a whole number. x is plus or minus 2. And I've already excluded those values. So I don't really have any new forbidden values from this. However, I know that because x squared minus 4 has roots x equals plus or minus 2, that does give me this useful piece of information. x squared minus 4 is the product x minus 2, x plus 2. So I can factor that. And I should be able to use this information to rewrite our equation this way. Factored form is actually a nice form if we can get it. So now I can multiply. And so the thing to notice here is if I multiply it by x minus 2 times x plus 2, that will eliminate this denominator entirely. That'll also eliminate this denominator and this denominator. So I can multiply by x minus 2 times x plus 2. And that gets rid of all of our denominators. So I'll multiply through by that. I'll do the cancellation that I can. x plus 2 cancels here. x minus 2 cancels here. And everything cancels here. And after all of that settles, I have 3 x minus 2 minus 3 x plus 2 equals x. And I'll expand and simplify. And there's a potential solution. It's not one of the forbidden values, so I'm good on that. And I should check to make sure that x equals negative 12 is in fact a solution. And they'll verify that if I drop x equals negative 12 into here, I ended with 3 over negative 10, 3 over 14. And this is actually a true statement. So x equals negative 12 is going to be our solution.