If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:11:49

Video transcript

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 you're probably right but let's start with the first problem we have X minus 1 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 any number divided by any other number and I say that they're going to be greater than zero 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 will definitely be true right we have a positive divided by a positive it'll definitely be a positive it'll definitely be greater than 0 or 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'll 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 and X plus 2 greater than 0 that's equivalent to this the top and the bottom are both 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 zero and X plus two less than zero if both of these are less than zero then you have a negative divided by negative which will be positive which will be greater than zero so let's actually solve in both of these circumstances so X minus 1 greater than 0 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 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 or whatever you want to think of it in order for both of these to hold true X has to be greater than 1 and X has to be greater than minus 2 right this statement we figured out means that X has to be greater than 1 and this statement so 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 right I mean 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 so the whole 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 right any number greater than 1 is definitely greater than negative 2 so that's one situation in which this equation holds true and we could 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 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 zero if we subtract two from both sides we get X is less than minus two 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 one and X has to be less than minus two now if something has to be less than one and it has to be less than minus two well it just has to be less than minus two anything less than minus two is going to be is going to satisfy both of these so this boils down to just X could also be less than minus two and remember this is an or either both the numerator and the denominator are positive or or they're both negative so both of them being positive boiled down to X could be greater than one or both of them being negative boils down to X is less than minus two so our solution is X could be greater than one or right X could be greater than one that's both of these positive or X is less than minus two 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 zero and then we have one so X could be greater than one not greater than or equal to so we put a little circle there to show that we're not including one and we could everything greater than one will satisfy this equation or anything less than minus two so we have minus one minus two anything less than minus two will also satisfy this equation by baking both the numerator and the denominator by both making the numerator and the denominator negative you could try it out minus three minus three minus one let's do minus three minus one is equal to minus 4 and then minus three plus two minus three plus two is equal to minus one minus four divided by minus one is positive four so all of these negative numbers also work now I told you that I will show you two ways of doing this problem so let me show you another way if you this one may be 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 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 what I'm going to no 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 so let's say that 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 could 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 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 what whether X plus 2 is positive or negative so let's do both situations so let's say if 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 X plus 2 is less than 0 right these are the two possibilities for X plus 2 actually in no situation can X plus 2 equals 0 if X plus 2 were to be equal to 0 then this whole expression would be undefined and so that definitely won't be a situation that we want to deal with it would be 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 right 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 could 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 as positive because X is greater than minus to multiply both sides by X plus 2 these cancel out 0 times X plus 2 is just 0 and then you have X minus 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 right or you could say if X is great well you could go both ways in that but we say look if 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 this for this branch we come up with a 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 and 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 saying 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 right 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 has to be less than minus two that'll satisfy this equation so our final result is X is either going to be greater than one or X is going to be less than minus two so once again you can graph it on the number line X is greater than one X is greater than one right there you have zero minus one minus two and then you have X is less than minus two you're not including the minus two and just like that and that is the exact same result we got up here so whatever 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