 This video will deal with rational equations. Our main goals when we're solving equations is to get rid of the fractions and we're going to do this by using the least common denominator and multiplying every piece by that whole least common denominator. And we need to look for any values that might be undefined by it being zero in the bottom. We'll have to check for those because if our answers end up being those then we can't include those answers. So when we solve rational equations, find any undefined values by setting each function and then denominator equal to zero. So take care of the domain right away. Then you find the least common denominator and you multiply each piece by the least common denominator and you want the whole thing. The whole least common denominator gets multiplied by each one to cancel out the denominators and then you can just simplify the either linear or quadratic equation that results and then you solve and check to make sure you don't have answers that are not in your domain. So we start here. What values make the denominators or denominators zero? Well that would be the x plus one can't be zero. So x being negative one would make the denominator zero. The least common denominator in this case is just x plus one because we only have one different denominator. So we want to multiply this term by x plus one and this term by x plus one. Because these are really like x plus ones over one. So I have a numerator x plus one and a denominator x plus one that cancel out and leave me with just the x plus five. And same thing here. I have a numerator x plus one that cancels a denominator x plus one and leaves me just with that eight. Now it's a simple linear equation and I subtract my five and find out that x is equal to three and I check and x only cannot be negative one so it can be three. So we're done. What values make my denominator zero? x minus four cannot equal zero or it would be undefined. So x cannot equal positive four. In the least common denominator, listing all my different denominators, all my denominators are the same so I just have x minus four. So I have to multiply by x minus four. And some people don't multiply, write it out like I do. They think it's a waste of time but I'm a scribbler. So I would write three times x minus four but since I wrote that x minus four above being multiplied by this fraction here, I can see that this one in the denominator cancels that one in the numerator and it's very easy to just see that I have minus five left. Equal to x minus four cancels x minus four and left with ten. So simplifying, three x minus 12 minus five is equal to ten. And three x minus 17 is equal to ten. Three x then is equal to 27 when I add the 17 to both sides. And when I divide by three, I find out that x is equal to nine. And again, the only value x cannot be is four so I can take my x equal nine. In this example, we can look at our denominators to find those values that are going to make them zero. And we have x minus one and that can't be equal to zero. And if that weren't equal to zero, adding one to both sides, then we find out that x can't be positive one. And then we have this different looking factor, which is one minus x can't equal zero. And if we bring the x to the other side, we find out that one is equal to or cannot equal x. Well, that's exact same domain value that restriction that I had from the other one. So these are very similarly related and we talked about these earlier when we talked about rational functions. We can actually factor this one as a negative, take the negative out. That makes that a negative one plus x. Or if you don't like that, you could write it as a negative and then a positive x minus one, they're the same thing. Times are the same on both the terms. And now you can see that we have many denominators with x minus one in them. So x minus one is definitely part of our least common denominator. But we also have to consider this negative. This one has x minus one, this one has x minus one, so does this. But we also have to consider that negative. So it's negative one times x minus one, that is our least common denominator. And when I come back and multiply everything by my least common denominator, then I can look and see what's going to happen here. Remember every term on both sides get multiplied by that least common denominator. And the x minus one cancels the x minus one leaving me with a negative times the three equal to and the negative, whichever one of these you want to take care of cancels the negative and the x minus one cancels the x minus one. So we really don't have anything in that denominator that we have to multiply by. So we just have five and then we have the x minus one that cancels the x minus one but we still have plus a negative x. So if we're continuing to solve, this is just a linear equation. And if I subtract the five from both sides, I'm going to get a negative eight equal to negative x. But remember we want positive x, so we have to divide by negative one. And x will be equal to a positive eight. And I look at my domain value and it's not one. So I can say that x equal eight. All right, what's going on here? We're going to have to factor this one. But my guess is these two. That happens a lot, but we'll make sure. I need factors of positive two that add up to negative three. So that means that both my signs are negative. And the only factors of two I have are two and one. So sure enough, they're the same. So all the different factors I have here, x minus one and x minus two. And neither one of them can be equal to zero. This one would tell me that x can't be one and this can't be equal to zero. So that would tell me that x can't be two. So there's my domain. And now I have to multiply everything by that whole least common denominator. x minus one, x minus two. In this first fraction or first term, x minus one's cancel. And I have three x being multiplied by the x minus two. Plus, and then in this one, the x minus two's are going to cancel. So I have my numerator of four and my numerator I multiplied by, that was x minus one. And that's equal to, and then x minus two cancels x minus two and x minus one cancels x minus one. So I'm just left with four. Then cancel down that one. Now I'm ready to distribute and try to solve. So I'm going to have three x squared here, minus six x. And then distributing again, the four times the x is four x minus four equal to four. Simplifying, we have three x squared minus six x plus four x will be minus two x. And then I have minus four equal four. But I need to bring this forward to the other side so I can, it's a quadratic. So I need to set it equal to zero. So three x squared minus two x and then minus eight equals zero. And I want to show you a neat little trick to factor that one. If you have the quadratic formula in your calculator, you can hit the program key. Now mine's a little bit different than a lot of these quadratic fours because I had to write it in. So I did it the shorthand, but here we go. Press enter and you should know how to use a quadratic formula. So I put in my a which is three and my b which is negative two and my c which is negative eight. And then mine just gives me the answers. So I can see that I have x equal two and then I had x equal some yucky thing. But if I take this fact, it actually came from the factors. If you remember we factored quadratics in all quadratic equations and we took our factors equal to zero and came up with something like this. So what I really want to do is take the x, the two back over there with that x so that I can see what factor it was. So I'm going to subtract two from both sides and have x minus two is equal to zero. So that's one of my factors. To be able to find the other factor, remember that the first term is made up of the factors of the first terms. Well I know I have x so I need to multiply by three x to make it three x squared. And I know that my last terms in my binomials come from the last term. Well I know I have negative two but if I multiply that by positive four, I'll have negative eight and now I know what my two factors are. So you can use a quadratic formula to help you factor. And now I just need to set these equal to zero so I can find out what x actually is. And if x minus two were equal to zero, x would be equal to two. And if three x plus four were equal to zero, then I'd have three x equal to negative four and that would tell me that I had x equal to negative four thirds. Now I have one more step. I have to go and check x equal to, oh, x can't equal to. So I cannot have this as one of my answers. So the only answer I have, negative four thirds is not one of my restrictions. So I only have that as my answer. So sometimes you get an answer that doesn't fit the domain so we have to exclude it. Now another way, starting over with this problem again, another way we could solve these is to set it equal to zero that makes it much easier to look at in the graph. So if I set this equal to zero, I actually would be subtracting the four to take it to the other side. I'd have to subtract this fraction and then I'd have zero on the other side. And now we bring our calculator up. I'm going to y equal and we enter in what we see. But remembering that whenever I have more than one term in a numerator and or denominator, I have to put it in parentheses. So again, I would have 3x divided by and then in parentheses, x minus one, close the parentheses, plus four divided by in parentheses, x minus two, minus the four divided by x squared minus 3x plus two that I brought over. And notice I haven't factored anything yet. I'm just going to put the left hand side on y one and the right hand side on y two. I think a standard window will probably work and I'm going to find where these graphs cross each other. Now remember that y equals zero is the x-axis. So right here is where my solution is going to be. Second trace five will help me find that. And I'm on the right part of my graph so I just press enter, enter, enter. And I see that x is equal to, that was our ugly decimal, negative 1.33 repeating. Notice, look at the graph again, that over here at two, which was my other answer, there is no answer. There's no intersection right here. So the nice thing about graphing is it'll only show you the answers that really work. It won't show you the ones that we would have to disregard if we did it on paper. The other advantage is that we don't have to factor anything. But remember, you have to be careful about your parentheses. That's probably the hardest part about graphing.