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.

well done explanation!

this guy is a idiot!!!!!!

@Yagyiu No, You are a idiot.

You need and English 101 version of KhanAcademy!

cant get any more boring…

thx dude

Dumb ass. Why don't you go fuck yourself and go somewhere else?

Thnx

Can someone tell me the rules for a closed dot and an open dot on the number line when graphing a quadratic inequality? Actually, I mean when you have one closed and one open.

hence "use any way you want"………………

Once again, Thank you so much.

thank you so much!!

Explanation is v. clear really

My textbook did that first method…and this makes it make sense:-)

your really good and i understand it could you make a part 3 if thats ok? and if u already have THANKS! i will check it out you are a really good taecher.

people that hate u are stupid and are not patiennt :/

NEVER MIND THOSE HATE COMMENTS 🙂

First method is easier for me 🙂

why can't I do this x-1>o*x+2 x-1>0 ???

What if the problem is greater than/less than or equal to? I remember that when you graph it on a number line either the variables in the numerator/denominator have to be open circles but I dont remember which.

I'm so glad you make these videos. I get so confused with the explanations the math teacher I have gives, but when you explain how to solve them, it just make sense. Thank you so much!

7:03 Aghhh nunununu

thanks bro

great work….!!!!

Wouldn't the first method of solving it be easier though?

lol this is exactly the problem I was having trouble with. Thank you.

very useful for me as i am taking math at college. Thank you very much Mr. Khan.

Where did you get the -3?

I dont know why, my teacher gives us weird problems. None of these work for the problems she gave us.

rational functions and inequalities, my most hated type of function. thanks for the video!

lol, this problem was #3 on my HW.

GOD BLESS YOU BECAUSE I HAVE A TEST IN PRE CALC TOMORROW AND THIS HELPED THANK YOUUUUU

What happens if x-1/x+2 > -x?

fuck this is hard

okay this is probably a very stupid question but I have this problem which is (3x-6)/(2x+1) + x > 0 what do I do with that +x? Like I'm completely lost .-.

bruh all you had to do was split each equation and do X-1=0 and X+2=0 and you'll get your two answers. easy.

amazin im still confused

You're taking too much time on explaining while you can just say that when you solve the x-1 and x+2 the signs of both -1 and +2 will changed into +1 and -2 lol

@[email protected]

You just saved my 1st Sem. Thank you. ^_^

Great !! Helped a lot…

WHAT ARE YOU TRYING TO IMPLY HERE? WHAT I WANT TO SEE HERE IS THAT YOUR FIRST STEP MUST BE EQUATING THE NUMERATOR AND DENOMINATOR TO ZERO. THANKS.

Why doesnt Khan Academy use the test point method?

Great explaination.

the reason why we can't cross multiply is because we are unsure if the unknown is negative

great

I’m still confused 🤷♂️

THANK YOU! You teach math 1000x better than my teacher

still confused ><

THE LESS AND GREATER THAN SIGNS THO now im confused, i was taught the opposite i think..

But -2 ISN'T >0….

U have 2 visualize the problem

Does that rule also apply to when a number is greater or equal to

Thanks

What of if it greater or equal to 2 or 1

I just set them to equal 0. Then slove as you would. Makes it a lot more easier

two ways to solve it? you trying too hard come on theres no difference between them you just making the "second step" more complicated its the same thing as step one

This is hard i dident get x+2 but i got the rest

wheres the graph? i dont want a single line graph. no one helps solve this graphically.

i understand how to solve rational functions but not solving rational inequalities. its a bit harder.

Really helpful

his "or" is very irrelevant. It stated that > why would he change it into <?

the organic chemistry tutor taught better than this channel…

I still don’t get it