Rational Equations - Video Tutorials & Practice Problems

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Introduction to Rational Equations

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Hey, everyone, we've solved a bunch of different linear equations. But now we're going to look at another type of equation called a rational equation. Now, our goal here is still the same. We want to find some value for X that will make our equation true. But you might be looking at this equation and not really be sure where to start. But don't worry, we're gonna take this rational equation and we're gonna turn it into a linear equation which we already know to solve. So let's get started. Now, a rational equation is an equation that has a variable like X in the denominator or the bottom of a fraction. So looking at my example here, I have one over X minus one equals 12. So this X that I have in my denominator tells me that I'm dealing with a rational equation. Now, like I said before, we can solve our rational equation by actually turning it into a linear equation. Now, something that's really important about rational equations is that our solution. So whatever value we get for X cannot be any value that's going to make a denominator zero. So in my example here this X minus one, this can never be equal to zero. Now, if I were to have solved this example, and I got the solution X equals one. And then I went to plug that back in, I would get 1/1 minus one, which would be one over zero. This is definitely not a fraction that I want to be dealing with because that denominator is zero. I want to avoid this at all costs. So this actually tells me that my denominator, that my answer, my solution cannot be one. So the fact that X cannot be one here is what's called the restriction. So I know that that word sounds a little bit complicated, but it just means that X cannot be any value that makes our denominator zero. So let's go ahead and jump into an example and look at the steps we need to take to solve a rational equation. So here I have X over X minus one is equal to 76. Now, my very first step here is going to be to determine my restriction. So we're actually going to determine what X can't be before we determine what X is. And we're going to do this by setting our denominator equal to zero. Now, we only need to do this with denominators that have a variable in them because we know that X or six is already not equal to zero. So we only need to do this with our denominators with variables. So I'm gonna take this X minus one and I'm gonna set it equal to zero. If I solve that, I'm going to add one to both sides that will cancel, I'm left with X equals one. So this tells me that if X was equal to one that would make my denominator zero, that's exactly what I don't want. So this is my restriction. So X cannot be equal to one. So we've completed step one, we've determined what X can't be. Let's go ahead and figure out what X is by taking our other steps. OK. So our step, our second step is going to be to multiply by our L CD, our least common denominator in order to get rid of those fractions. This is exactly what's going to take us back to a linear equation, which is really familiar to us. So looking at my denominators over here, I have X minus one and six. And since these two denominators don't have any common multiples that tells me that my least common denominator is just the product of these. So to eliminate these fractions, I'm going to multiply by six times X minus one. That is my least common denominator. Remember when we're eliminating fractions. When we're multiplying by our L CD, we need to make sure and distribute it to every single term in our equation. So let's go ahead and expand this. So I get six times X minus one times my first term over X minus one and that's equal to six times X minus one. My L CD again, times my second term 7/6. Now let's go ahead and simplify this a little bit. So on my left side here, my X minus one is going to cancel, which is great because I am just left with six X and that is equal to on this side, my sixes are going to cancel. So I'm left with seven times X minus one. I just moved that seven over to the front just to make it a little bit easier to deal with. So we've completed step two and you might have noticed that now we just have a linear equation. So our third step is simply to solve that linear equation. So the first thing I need to do here is to distribute this seven into my parentheses. My left side is going to remain the same. So I'm just left with six X equals this distributed seven is going to give me seven X minus seven. I don't have any like terms on either side that need to get combined. So I can go ahead and move all my X terms to one side, all my constants to the other. I want to move this seven over here since my constant is already on that right side. So to do that, I need to subtract seven X from both sides, it will then cancel and I am left with a negative one X is equal to negative seven. Now, my very last step in solving this linear equation is going to be to isolate X which I can do by dividing by negative one, my negative one will cancel and I am left with X equals negative seven divided by negative one. It gives me positive seven and this is my solution. So step three is done but we still have a final step. Step four, which is going to be to check our solution with our restriction. So step one, we found our restriction, we're gonna check our solution and make sure that it's not that. So my restriction here was that X cannot be equal to one. And for my solution, I got X equals seven which is definitely not one. So we're good to go and our solution is X equals seven. That's all for this one. I'll see you in the next one.

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Restrictions on Rational Equations

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Hey, everyone, when solving a bunch of rational equations, you may end up getting a solution that's equal to your restriction. So you may do step one and determine that your restriction is X cannot be equal to one. You move on to step two and step three. And you find that your solution is X equals one. Now, when you go to step four to check your solution with your restriction, you find that your solution is exactly what you said that X can't be. So I'm gonna show you exactly what to do when that happens. Let's just jump right into an example. So I have X minus five over X minus two equals negative three over X minus two plus six. Now, my very first step here is of course to determine and my restriction. So both of my denominate denominators here, I have X minus two. So what would X be to make those denominators zero? Well, if I plugged two and for each of those X's, that would make my denominator zero, which is what I don't want. So I know that X cannot be equal to two and that is my restriction. So I've completed. Step number one. Now, step number two is to multiply by our least common denominator to eliminate our fractions. So again, both of my denominators here are X minus two, which tells me that my least common denominator is also X minus two. So remember when we multiply by our least common denominator, we want to distribute it to every single term in our equation. Make sure you don't forget that six over there, it might be a little bit hard to see. So let's go ahead and expand this out. So I get X minus two times X minus five over X minus two equals X minus two. My least common denominator multiplied by this term negative three over X minus two plus my very last term X minus two times six. Now, this looks a little bit crazy now, but remember we're going to simplify it down a ton to end up with a linear equation. So, on my left side, I see that my X minus two is canceled there and I'm just left with X minus five. Now, that's equal to my right side which again, I see I have an X minus two that will cancel leaving me with a negative three. And then my last term, I'm gonna go ahead and move that six to the front of my parentheses to make it a little easier to work with six times X minus two. And I have completed step number two. Now we can simply continue solving this as though it's a linear equation because it is. So let's go ahead and simplify it a little bit more. We need to go ahead and distribute that six into the parentheses. There's nothing else I need to distribute. So the left side of my equation is gonna stay the same X minus five equals negative three and then my distributed 66 X minus 12. Now I need to go ahead and combine all of my like terms again, on my left side, I don't have anything to do here. I can just simply leave it as X minus five equals. I do have some like terms over on my right side, I have this negative three and this negative 12 which are gonna go ahead and combine to give me negative 15. Then I need to bring my positive six X down and then I need to go ahead and complete the step in my linear equation solving which is moving all of my X terms to one side, all of my constants to the other. So I want to go ahead and move this 15 over to the left side and move my X over to the right side. So in order to do that, I need to go ahead and add my 15 to both sides. Of course, because whatever I do to one side, I have to do to the other and then subtract my X from both sides as Well, so I am left now with negative five plus 15 which gives me positive 10 equals on said I have six X minus X which gives me five X. Now my very last step is to isolate X which here I can do by dividing both sides by five. And I am left now with two equals X because those fives cancel out. So if I want to go ahead and flip my solution around, just to put it in this form, I get X equals two. So I've completed step number three, I have my solution. I'm done solving. But I need to go back to step number four and go ahead and check my solution with my restriction. So my restriction here says that X cannot be equal to two. But I literally just got the solution that X is equal two. So what does that even mean? Well, if I get an answer that's equal to my restriction, then there is no solution at all. So the fact that I got this X equals to doesn't matter because I have no solution. If my solution is equal to my restriction, X cannot be that number. So it can't be my solution. I have none. Now another way to say this is to say that our solution set is equal to our empty set because there is no number that is going to solve this rational equation. So any time you get a solution and you find that it's equal to your restriction. You have no solution at all. That's all for this one. Thanks for watching.

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Problem

Problem

Solve the equation. $\frac{2x+4}{x-1}=5$

A

$x=3$

B

$x=1$

C

$x=5$

D

No solution

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Problem

Problem

Solve the equation. $\frac{5}{x}-\frac{2}{3x}=4+\frac{3}{x}$

A

$x=0$

B

$x=1$

C

$x=\frac13$

D

No solution

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Problem

Problem

Solve the equation. $\frac{-5}{x+4}-3=\frac{x-1}{x+4}$