Algebra

Rational inequalities | Polynomial and rational functions | Algebra II | Khan Academy

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In this video I want to solve some problems with inequalities, which are quite intricate. You might say, “Aren\’t all problems inequal? tangled? ” To some extent, you’re probably right. But let’s start with the first task. We have x minus 1 over x plus 2 is greater than 0. In fact, I’m going to show you two ways to solve it. The first way I think is a bit simpler. But I will show you both methods to judge which one it’s easier for you. The first way I can think of is if we just have one number divided by another number and tell you that will be greater than 0. We just have to remember the properties of multiplication and division of negative numbers. In what situation this liver will be greater than 0? It will be greater than 0 only if both numbers are – so we can write that a and b are both, greater than 0. This is one of the circumstances in which this it will definitely be true. We have the positive divided by the positive; this will definitely be a positive number. It will definitely be greater than 0. Or we may have a situation where we have a negative divided by negative. If we have one sign divided by the same sign, this will also be a positive number. So we have “or” a is less than 0 and b is less than 0. When you have some kind of rational expression like the one greater than 0 has two situations, in which this inequality will be true. The numerator and denominator are both greater than 0 or both are less than 0. Let’s remember it and actually solve this problem. There are two situations to solve the problem. The first is when both are greater than 0. If this and that are greater than 0, we will be fine. We can say that in the first situation – I will draw small tree – we have x minus 1 is greater than 0 and x plus 2 is greater than 0. This is equivalent to that. Upper and lower – if both are greater than 0. then when we divide them we will get something greater than 0. The other option we saw a while ago is if both are less than 0. The other option is x minus 1 is less than 0 and x plus 2 is less than 0. If both are less than 0, then we will have a negative divided by a negative, which will be a positive number. Which will be greater than 0. Let’s actually solve the inequality in both circumstances. x minus 1 is greater than 0. If we add 1 to both sides, we get x is greater than 1. If we solve x plus 2 is greater than 0, then if we subtract 2 from both sides of the equation – remember I’m doing this now – we get x is greater than minus 2. So to be both true – with this slightly brown or red as you see it – to be both true, x must be greater than 1 and x must be greater than minus 2. We found that this statement means that x must be greater of 1, and this statement shows us that x should be greater than minus 2. If x is greater than 1 and x must be greater than minus 2, it is clear that x must be greater than 1. 0 will not satisfy the inequality because it is greater from minus 2, but not greater than 1. To be something greater than 1 and minus 2, it must be greater than 1. In this whole line of thought, where I say the numerator and the denominator is greater than 0, this will only happen, if x is greater than 1. Because if x is greater than 1, then it will definitely be greater than minus 2. Each number after 1 is definitely larger from minus 2. So this is a situation where the equation is true, as we can even check it out. Let’s say x is 2. 2 minus 1 is 1 over 2 plus 2. This is 1/4. We have a positive number. Now let’s look at a situation in which both are negative. If x minus 1 is less than 0, if we add 1 to both sides of this equation, this shows us that x minus 1 is less than 0. It’s the same – if we add 1 to both sides – as if to say that x is less than 1. So this condition comes down to this condition. This condition, x plus 2 is less than 0. If we subtract 2 from both sides, we get x is less than minus 2. Therefore, this condition is reduced to this condition. So to be both negative, the numerator and denominator must be a negative number – we know that x must be less than 1, and x must be less than minus 2. If something should be less than 1 and should be less from minus 2, it must only be less than minus 2. Anything less than minus 2 will satisfy both conditions. So it just comes down to x can be too less than minus 2. And remember, it\’s an “or.” Or the numerator and denominator are positive, or both are negative. Both being positive is reduced to x may be greater from 1 or both to negative is reduced to x is less than minus 2. So the solution is x can be greater than 1 or – this is if both are positive – or x is less than minus 2. This is if both are negative. If we wanted to draw it on a numerical axis – let me draw a numerical axis like this. This can be 0 and then we have 1, so x may be greater than 1. Not greater than or equal to. So we put a little circle in there to show that we do not include 1. Anything greater than 1 will satisfy this equation. Or anything less than minus 2. We have minus 1, minus 2, everything less than minus 2 will also satisfy the equation by making the numerator and the denominator negative. As you can check it out. Minus 3. Minus 3, minus 1. I’ll just calculate it, minus 3, minus 1 is equal to minus 4. Then minus 3 plus 2. Minus 3 plus 2 is equal to minus 1. Minus 4 divided by minus 1 is plus 4. So all these negative numbers will also work. I told you I would show you two ways to solve this problem. Let me show you another way, if you like it a little confusing. The other way – let me rewrite the task. you have x minus 1 over x plus 2 is greater than 0. In fact, let’s mix them up a bit and you can to apply the same logic. Let’s say it’s greater than or equal to – not really. I will leave it the same way, and maybe in the next video I will use the case where we have greater than or equal to, because I really don’t want to – maybe I will to increase the level of difficulty by one step. We just say that x minus 1 over x plus 2 is simple greater than 0. One of the things you might say is that if I have a rational expression like this, maybe multiply both sides of the equation on x plus 2. So I will get rid of it in the denominator and I can multiply it by 0 and get it out of my way. But the task is if you multiply both sides of an inequality by a given number – if you multiply by positively, you keep the inequality the same. But if you multiply by the negative, you have to change the sign of the inequality, not knowing if x plus 2 is positive or negative. So let’s do it in both situations. Let’s use the situation where x plus 2 – I will write it this way. x plus 2 is greater than 0. Then the other situation is when – let it write in a different color. Where x plus 2 is less than 0. There are two options for x plus 2. In fact, can x plus 2 be equal to 0 in these situations? If x plus 2 must be equal to 0, then this whole expression will be indefinite. So this is definitely not going to be a situation that we want to get. This will be an uncertain situation. These are the two situations in which we multiply both sides. If x plus 2 is greater than 0, it means that x is greater than minus 2. We can subtract 2 from both sides of this equation. If x is greater than minus 2, then x is plus 2 is greater than 0. Then we can multiply both sides of this equation for x plus 2. We have x minus 1 over x plus 2 is greater than 0. I’m going to multiply both sides by x plus 2, which assume it is positive because x is greater than minus 2. I multiply both sides by x plus 2. These are shortened. 0 over x plus 2 is only 0. You are left with only x minus 1 is greater than – this is simplified to 0. Solve it for x, add 1 to both sides and get x is greater than 1. We saw that if x plus 2 is greater than 0 or we could let’s say if x is greater than minus 2, then x too must be greater than 1. Or you can say that if x is greater – you can to do it both ways. But we say that both must be true. If x satisfies both, then simply must be greater than 1. Because if it’s greater than 1, then definitely will satisfy this condition here. We get the solution for this branch x is greater than 1. This is one situation where x plus 2 is greater than 0. The other situation is x plus 2 is less than 0. If x plus 2 is less than 0, this is equivalent to x is less than minus 2. You just subtract 2 from both sides. If x plus 2 is less than 0, what should we do, when we multiply both sides – let’s do it. We have x minus 1 over x plus 2. We have some inequality and then we have 0. If we multiply both sides by x plus 2, x plus 2 is a negative number. When we multiply both sides of an equation by negative number, we need to change the sign of the inequality. Therefore, the sign for more will become a sign for less, because we assume that x plus 2 is negative. These are shortened. 0 in all is 0. We get x minus 1 is less than 0. Solve it for x, add 1 to both sides, x is less than 1. In the case where x plus 2 is less than 0 or x is less from minus 2, x must be less than 1. I mean, we know that if you say something has to be less from minus 2 and less than 1, you just say it’s less from minus 2. Anything less than minus 2 will satisfy this condition. But not everything that satisfies this will satisfies this as well. This is the only condition that interests us. In the case where x plus 2 is less than 0, we just can let’s say x must be less than minus 2. This will satisfy this equation. The end result is x will be either greater of 1, or x will be less than minus 2. Again, I can draw it on the number axis. x is greater than 1 there. You have 0, minus 1, minus 2. Then you have x is less than minus 2, no we include minus 2. Here’s how. As this is exactly the same result we got up here. So you can use the option you find easier. But you see that both are a bit nuanced, and you should to think a little about what happens when you multiply or divide by positive or negative numbers.

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