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Rational inequalities: both sides are not zero

Sal solves the rational inequality (x-3)/(x+4)≥2. Created by Sal Khan.
Video transcript
Let's tackle a slightly harder problem than what we saw in the last video. I have here x minus 3 over x plus 4 is greater than or equal to 2. So the reason why this is slightly harder is I now have a greater than or equal to. And the other thing that makes it slightly harder is I don't just have a simple 0 here, I actually have a 2 here. So I'm going to do this problem the same way. I'm going to do it two ways. I'm going to do the same two ways we did last time, but I'm going to reverse the order of the two ways that I do it. So the first thing I can do, I might be tempted-- let me multiply both sides of this equation times x plus 4. And like we saw in the last video we had to be very careful because we have an inequality here. If x plus 4 is greater than 0 then we're going to keep the inequality the same. If it's less than 0 we're going to swap it. So let's look at those two situations. So one situation is-- let me make it right here. One situation is x plus 4 is greater then 0. And remember, it can ever be equal to 0 because then this expression would be undefined. So let's explore x plus 4 is greater than 0. If we subtract 4 from both sides of this equation this is equivalent to saying-- these are equivalent statements. If we subtract 4 from both sides that x is greater than minus 4. So if we assume that x is greater than minus 4 then x plus 4 is going to be greater than 0. And then when we multiply both sides of this times x plus 4-- let's do that. x minus 3 over x plus 4. You have your 2. We're going to multiply both sides of this times x plus 4. Since we're assuming x is greater than minus 4 or that x plus 4 is greater than 0, we don't have to switch the inequality sign. We're multiplying both sides by a positive. So the inequality stays the same as in our original problem. And the whole reason why we did that is because this and this will cancel out. And then we have 2 times x plus 4. Let's see what we get. x minus 3 is greater than or equal to 2 times x plus 2 times 4, which is 8. And now what can we do? We can subtract x from both sides. We get minus 3 is greater than or equal to x plus 8. I just took x from 2x. And then we can subtract 8 from both sides. So you subtract 8 from both sides. You get minus 3 minus 8 is minus 11 is greater than or equal to x. And of course, then we subtract an 8 so this guy disappears. So if we assume that x is greater than minus 4, then x has to be less than or equal to minus 11. Now that seems a little bit nonsensical. In order for this statement to be true, x has to be both greater than minus 4 and it has to be less than minus 11. Anything greater than minus 4 is going to be greater than minus 11. So there's no x that satisfies this equation. So if we assume that x plus 4 is greater than 0, we just end up with a nonsensical solution. So we can actually ignore that path. There's no situation in which x plus 4 is going to be greater than 0 and we're going to actually have a solution to this equation. So let's assume what happens when x plus 4 is less than 0. I'll do it here. So what happens when x plus 4 is less than 0? And once again, that's an equivalent statement. If we subtract 4 from both sides of this equation to saying that x is less than minus 4. These are equivalent statements. So if we assume that x is less than minus 4, then when we multiply both sides of this equation times x plus 4 and you're multiplying it by x minus 3 over x plus 4. And you have your 2 and you're multiplying that times x plus 4. Since we're taking the other situation where x is less than minus 4, which makes this expression negative-- since we're multiplying both sides of the equation by a negative, we swap the inequality just like that. And let's see what we get. This and this cancel out. And then we get x minus 3 is less than or equal to 2 times x plus 2 times 4, which is 8. And we do the same algebra. Let's subtract x from both sides. We get minus 3 is less than or equal to x plus 8. If we subtract 8 from both sides we get minus 11 is less than or equal to x. If we assume this is negative, we get x has to be less than minus 4. If x is less than minus 4, in order for this equation to hold up, x also has to be greater than minus 11. And this is actually possible. You can have an x, so x has to be-- we can rewrite this-- greater than or equal to minus 11. I just swapped the two sides. And x has to be less than minus 4. And there are x's that satisfy these two constraints. There weren't x's that satisfy these two constraints. Nothing satisfies both of these, but there are things that satisfy both of these. For example, minus 5. Let's draw this on the number line. Let's draw a solution to this problem. Let me keep the problem there. If I draw a number line, I want to switch colors. So we have minus 11. We can be greater than or equal to minus 11. So this is minus 11. And then we have minus 4 over here. Maybe 0 sitting over there. We have to be less than minus 4. Remember, we can ever be equal to minus 4 because that would make this undefined. So we have to be less than minus 4, so we don't include minus 4. We're less than minus 4, but we have to be greater than or equal to 11. So we can't go less than 11, but we include the 11 because this is greater than or equal to. So anything in this, where I've shaded it in-- anything on this range will satisfy that equation over there. Now let's do it the other way. Let's do it where we-- the numerator and denominator both have to be positive or negative. Let's see if we can do the same problem that way. So let me rewrite it down here, maybe in a new color. So the same problem. x minus 3 over x plus 4 is greater than or equal to 2. Now, in order to do that reasoning that I did in the first video, remember the first video I said, well, it's a divided by b is greater than 0. Either both of these are positive or both of these are negative. But this only worked when we had a 0 over here. But we don't have a 0 here we have a 2 so we can't at least immediately do that rationale. But maybe we can if we subtract 2 from both sides of this equation. So let's do that. So you have x minus 3 over x plus 4 minus 2 is greater than or equal to 0. Just subtract a 2 from both sides. I can add or subtract without messing up the inequality. Never have to worry about that. It's only when you multiply by negatives that you swap the inequality. And now, what is this? You know, this doesn't look like a rational expression. But I can rewrite minus 2, right? Minus 2 is the same thing as minus 2 times x plus 4 over x plus 4. This is minus 2 times 1. This is the same thing. And this is the same thing as minus 2x plus-- let me be careful. This is the same thing as minus 2x minus 8 over x plus 4. This right here is the exact same thing as minus 2. I just multiply it times x plus 4 for x plus 4. Let's write it this way because then we can add them. So if we write that way, so I have x minus 3 over x plus 4. And I wrote it that way so I have a common denominator. And then I have plus minus 2x minus 8 over x plus 4 is greater than 0. I have a common denominator. That was my whole point of doing this. And so I have a denominator of x plus 4 and then when you add these two things you get x minus 2x. That's a minus x. And then you have minus 3 minus 8-- that's minus 11-- is greater than 0. So now I simplify the problem into something like this. So there's two situations. Oh, let me be very careful. This was greater than or equal. This must be greater than or equal. Don't want to lose that little equal sign. That was the whole point of this problem. So now I'm ready to operate just like this. Actually, before I even do this, let's get one thing out of the way. Let's be very clear. x can never be equal to minus 4. Because if x is equal to minus 4, this whole statement-- that original problem is undefined. So let's just write that first. You know, we could have written that from the get-go. x cannot be equal to minus 4, for any solution. That will make this expression undefined. Now, we have this and we saw in the first video and we saw right there that there's two situations. Either both of these are positive. So they're either both positive, so minus x minus 11 is greater than or equal to 0. This could be equal to 0. Actually, if this is equal to 0, then this whole thing is going to be true. So either that is greater than or equal to 0 and this right here is greater than 0. It can't be equal to 0. And x plus 4 is greater than 0. That's the situation where both of them are positive. Or they're both going to be less than 0. And I'm running out of space. Or minus x minus 11 is less than or equal to 0. And-- I'll write and like that. And x plus 4 is less than 0. Remember, it can't ever equal 0. So they're either both positive or both negative. That's the only way they we're going to get something. Well, the numerator can be equal to 0 because this thing can equal 0. Or they're both positive to get something positive. Or they're both negative, so that when you divide a negative by a negative you still get something positive. So let's look at these two situations and see if we get any solutions that make sense. So right here, if we add x to both sides of this equation we get minus 11 is greater than or equal to x. Or another way you could say it, is x is less than or equal to minus 11. I just swapped these two sides. And then here, if we add minus 4-- if we subtract 4 from both sides we get x is greater than minus 4. But once again, this makes no sense. There's no way that I can have-- remember there's an and here. Both of these have to be true. There's no x that is greater than minus 4 and less than minus 11. So this right here, there's no constraints. So both of these can never be-- both the numerator and the denominator can never both be positive at the same time because in order for them both to be positive you get an impossible constraint to fill. So let's look at the situation. The other situation is when they are both negative. When both this numerator and this denominator are both negative. Let me be clear. You can't make a statement about these two not being positive. These two things can ever be positive. Remember, because we had to get the 0 here in order to use this logic to begin with. So the fact that when we tried to figure out x's that would make both of these positive, we got a nonsensical, impossible to fill constraint, tells me that both this numerator and this denominator can never both be positive for any given x. Now let's see if they can both be negative. So for both of them to be negative or the numerator could also be equal to 0. If we add x to both sides of this we get minus 11 is less than or equal to x. Or we could say that x is greater than or equal to minus 11 just swapping around. Over here, if we subtract 4 from both sides we have x is less than minus 4. And-- and this is an and. So x has to be greater than or equal to minus 11 or less than minus 4, which is exactly what we came up in the last problem. I can draw the number line again. We have to be greater than or equal to minus 11. So that's minus 11 there. We could be equal to it or greater than, but we have to be less than minus 4. So this is minus 4. We can't even be equal to minus 4, that'll make our denominator undefined or our whole equation undefined. It will make our denominator 0. And so this is our solution set. Anything right here that I've shaded in, including minus 11, but not including minus 4. So if you could get this problem I think you can handle pretty much any rational equation problem. Hope you enjoyed that.