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Practice English Speaking&Listening with: Rational inequalities | Polynomial and rational functions | Algebra II | Khan Academy

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In this video I want to do a couple of inequality problems

that are deceptively tricky.

And you might be saying, hey, aren't all inequality

problems deceptively tricky?

And on some level you're probably right.

But let's start with the first problem.

We have x minus 1 over x plus 2 is greater than 0.

And I'm actually going to show you two ways to do this.

The first way is, I think, on some level, the simpler way.

But I'll show you both methods and whatever works for you,

well, it works for you.

So the first way you can think about this, if I have just any

number divided by any other number and I say that they're

going to be greater than 0.

Well, we just have to remember our properties of multiplying

and dividing negative numbers.

In what situation is this fraction going

to be greater than 0?

Well, this is going to be greater than 0 only if both a

and-- so we could write both a is greater than 0 and

b is greater than 0.

So this is one circumstance where this'll

definitely be true.

We have a positive divided by a positive; it'll

definitely be a positive.

It'll definitely be greater than 0.

Or we could have the situation where we have a negative

divided by a negative.

If we have the same sign divided by the same sign we're

also going to be positive.

So or-- a is less than 0 and b is less than 0.

So whenever you have any type of rational expression like

this being greater than 0, there's two situations in

which it will be true.

The numerator and the denominator are both

greater than 0, or they're both less than 0.

So let's remember that and actually do this problem.

So there's two situations to solve this problem.

The first is where both of them are greater than 0.

If that and that are both greater than 0, we're cool.

So we could say our first solution-- maybe I'll draw a

little tree like that-- is x minus 1 greater than 0 and

x plus 2 greater than 0.

That's equivalent to this.

The top and the bottom-- if they're both greater than 0

then when you divide them you're going to get

something greater than 0.

The other option-- we just saw that-- is if both of

them were less than 0.

So the other option is x minus 1 less than 0 and

x plus 2 less than 0.

If both of these are less than 0 then you have a negative

divided by a negative, which will be positive.

Which will be greater than 0.

So let's actually solve in both of these circumstances.

So x minus 1 greater than 0.

If we add 1 to both sides of that we get

x is greater than 1.

And if we do x plus 2 greater than 0, if we subtract 2 from

both sides of that equation-- remember I'm doing this

right now-- we get x is greater than minus 2.

So for both of these to hold true-- so in this little brown

or red color, whatever you want to think of it-- in order for

both of these to go hold true, x has to be greater than 1 and

x has to be greater than minus 2.

This statement we figured out means that x has to be greater

than 1 and this statement tells us that x has to

be greater than minus 2.

Now, if x is greater than 1 and x has to be greater than minus

2, x clearly has to be greater than 1.

You know, 0 would not satisfy this because 0 is greater

than minus 2, but it's not greater than 1.

So for something to be greater than 1 and minus 2 it has

to be greater than 1.

This whole chain of thought where I'm saying the numerator

and the denominator are greater than 0, that's only going to

happen if x is greater than 1.

Because if x is greater than 1, then x is definitely going to

be greater than negative 2.

Any number greater than 1 is definitely greater

than negative 2.

So that's one situation in which this equation holds true,

and we can even try it out.

Let's say x was 2.

2 minus 1 is 1 over 2 plus 2.

It's 1/4.

It's a positive number.

Now let's do the situation where both of these

are negative.

If the x minus 1 is less than 0, if we add 1 to both sides of

that equation that tells us so x minus 1 is less than 0.

That's the same thing-- if we add 1 to both sides of that--

as saying that x is less than 1.

So that constraint boils down to that constraint.

Now this constraint, x plus 2 is less than 0.

If we subtract 2 from both sides we get x

is less than minus 2.

So this constraint boils down to that constraint.

So in order for both of these guys to be negative, both the

numerator and the denominator to be negative-- we know that

x has to be less than 1 and x has to be less than minus 2.

Now if something has to be less than 1 and it has to be less

than minus 2, well, it just has to be less than minus 2.

Anything less than minus 2 is going to satisfy both of these.

So this boils down to just x could also be

less than minus 2.

And remember, this is an or.

Either both the numerator and the denominator are positive,

or they're both negative.

So both of them being positive boil down to x could be greater

than 1, or both of them being negative boils down to x

is less than minus 2.

So our solution is x could be greater than 1 or-- that's

both of these positive-- or x is less than minus 2.

That's both of these negative.

And if we wanted to draw it on a number line-- let me draw a

number line just like that.

That could be 0 and then we have 1, so x could

be greater than 1.

Not greater than or equal to.

So we put a little circle there to show that

we're not including 1.

And everything greater than 1 will satisfy this equation.

Or anything less than minus 2.

So we have minus 1, minus 2, anything less than minus 2 will

also satisfy this equation by both making the numerator

and denominator negative.

And you could try it out.

Minus 3.

Minus 3 minus 1.

I'll just do minus 3 minus 1 is equal to minus 4.

And then minus 3 plus 2.

Minus 3 plus 2 is equal to minus 1.

Minus 4 divided by minus 1 is positive 4.

So all of these negative numbers also work.

Now, I told you that I would show you two ways

of doing this problem.

So let me show you another way if you found this one maybe

a little bit confusing.

So the other way-- let me rewrite the problem.

You get x minus 1 over x plus 2 is greater than 0.

And actually, let's mix it up a little bit and you could

apply the same logic.

Let's say it's greater than or equal-- well, actually no.

I'll just keep it the same way and maybe in the next video

I'll do the case where it's greater than or equal just

because I really don't want to-- maybe I want to

incrementally step up the level of difficulty.

We're just saying x minus 1 over x plus 2 is just

straight up greater than 0.

Now one thing you might say is well, if I have a rational

expression like this, maybe I multiply both sides of this

equation by x plus 2.

So I get rid of it in the denominator and I can

multiply it times 0 and get it out of the way.

But the problem is when you multiply both sides of an

inequality by a number-- if you're multiplying by a

positive you can keep the inequality the same.

But if you're multiplying by a negative you have to switch the

inequality, and we don't know whether x plus 2 is

positive or negative.

So let's do both situations.

Let's do one situation where x plus 2-- let

me write it this way.

x plus 2 is greater than 0.

And then another situation where-- let me do that

in a different color.

Where x plus 2 is less than 0.

These are the two possibilities for x plus 2.

Actually, in those situations can x plus 2 equal 0?

If x plus 2 were to be equal to 0 than this whole expression

would be undefined.

And so that definitely won't be a situation that we

want to deal with it.

It would an undefined situation.

So these are our two situations when we're multiplying

both sides.

So if x plus 2 is greater than 0 that means that x

is greater than minus 2.

We can just subtract 2 from both sides of this equation.

So if x is greater than minus 2, then x plus

2 is greater than 0.

And then we can multiply both sides of this

equation times x plus 2.

So you have x minus 1 over x plus 2 greater than 0.

I'm going to multiply both sides by x plus 2, which I'm

assuming is positive because x is greater than minus 2.

Multiply both sides by x plus 2.

These cancel out.

0 times x plus 2 is it just 0.

You're just left with x minus 1 is greater than-- this

just simplified to 0.

Solve for x, add 1 to both sides, you get

x is greater than 1.

So we saw that if x plus 2 is greater than 0, or we could

say, if x is greater than minus 2, then x also has

to be greater than 1.

Or you could say if x is great-- well, you could

go both ways in that.

But we say, look, both of these things have to be true.

If for x to satisfy both of these, it just has

to be greater than 1.

Because if it's greater than 1 it's definitely

going to satisfy this constraint over here.

So for this branch we come up with the solution

x is greater than 1.

So this is one situation where x plus 2 is greater than 0.

The other situation is x plus 2 being less than 0.

If x plus 2 is less than 0 that's equivalent to saying

that x is less than minus 2.

You just subtract 2 from both sides.

Now, if x plus 2 is less than 0, what we'll have to do

when we multiply both sides-- let's do that.

We have x minus 1 over x plus 2.

We have some inequality and then we have a 0.

Now if we multiply both sides by x plus 2, x plus

2 is a negative number.

When you multiply both sides of an equation by a

negative number you have to swap the inequality.

So this greater than sign will become a less than sign because

we're assuming that the x plus 2 is negative.

These cancel out.

0 times anything is 0.

We get that x minus 1 is less than 0.

Solving for x, adding 1 to both sides, x is less than 1.

So in the case that x plus 2 is less than 0 or x is less than

minus 2, x must be less than 1.

Well, I mean we know-- if you say something has to be less

than minus 2 and less than 1, just saying that it's less

than minus 2 will do the job.

Anything less than minus 2 is going to satisfy this one.

But not anything that satisfies this one is going to

satisfy that one.

So this is the only constraint we have to worry about.

So in the event where x plus 2 is less than 0, we can just

say that x has to be less than minus 2.

That'll satisfy this equation.

So our final result is x is either going to be greater

than 1 or x is going to be less than minus 2.

So once again, I can graph it on the number line.

x is greater than 1 right there.

You have 0, minus 1, minus 2.

And then you have x is less than minus 2, you're not

including the minus 2.

And just like that.

And that is the exact same result we got up here.

So whichever version you find to be easier.

But you can see, they're both a little bit nuanced and you have

to think a little bit about what happens when you multiply

or divide by positive or negative numbers.

The Description of Rational inequalities | Polynomial and rational functions | Algebra II | Khan Academy