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# Rational inequalities: both sides are not zero

## Video transcript

let's tackle a slightly harder problem than what we saw in the last video I have here X minus three over X plus 4 is greater than or equal to two 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 let's 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 have 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 than 0 and remember it can never 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 statement 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 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 right 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 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 right I just took X from 2x and then we can subtract 8 from both sides so you should attract 8 from both sides you get minus 3 minus 8 is minus 11 is greater than or equal to X and of course if 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 a little bit nonsensical right if 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 a greater than minus 11 so there's no X that satisfies this equation right if so if we assume that X plus 4 is greater than 0 we just get up and we just end up with a nonsensical solution so we can actually ignore we can 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 maybe 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 just 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 so you have X plus 4 and you're multiplying it by X minus 3 over X plus 4 and all of that 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 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 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 so if we assume so this this is interest so if 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 swap 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 so let's draw this on the number line let's draw our solution to our to this problem and we keep the problem there and 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 we have minus 4 over here maybe 0 sitting over there we have to be less than minus 4 remember we can never 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 was 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 logic or they either have to be the numerator denominator both have to be positive or negative let's see if we can do that 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 greater than or equal to two now in order to do that reasoning that I did in the first video remember the first video I said well if a divided by B is greater than zero either both of these are positive or both of these are negative but this only worked when we had a zero over here but we don't have a zero here we have a two so we can't at least immediately do that rationale but maybe we can if we subtract two from both sides of this equation so let's do that so you have X minus three over X plus four minus 2 is greater than or equal to zero just subtracted two from both sides don't have you know 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 minus 2 is the same thing as minus 2 times X plus 4 over X plus 4 all right this is minus 2 times 1 this is the same thing and this is the same thing as minus 2 X plus let me be careful this is the same thing as minus 2 X minus 8 right over X plus 4 this right here is the exact same thing as minus 2 right I just multiply it times X plus 4 over X plus 4 let's write it this way because then we can add them so if we write it that way so I have X minus 3 over X plus 4 and I wrote it that way to 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 2 X that's minus X and then you have minus 3 minus 8 that's minus 11 is greater than 0 so now I've simplified the problem into something like this so there's two situations oh let me be very careful I lost the this was greater than or equal this must be greater than or equal this is greater than or equal don't want to lose little equal signer that was the whole point of this problem so now I'm ready to operate it on on just like this so let's 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 right 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 X minus X minus 11 is greater than or equal to 0 right this could be equal to 0 if 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 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 all right and like that and X plus 4 is less than 0 remember can't ever equal 0 so they're either both positive or both negative that's the only way that 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 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 right 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 X is greater than minus 4 but once again this makes it there's no way that I can have remember there's an end here both of these have to be true there is no X that is greater than minus 4 and less than minus 11 so this this right here there's no constraints that so both of these can never be both the numerator 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 this situation the other situation is when they are both negative when both this numerator and this denominator are both negative all right let me be clear these two I didn't you can't make a statement about these two not being positive these two things can never be positive remember because we had to get the zero here in order to use this logic to begin with so the fact that when we try 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 them around over here if we subtract 4 from both sides we have X is less than minus 4 and and this is an an 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 and we 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 will make our denominator undefined or our holy equation undefined it'll make our denominator 0 and so this is our solution set anything right here that I have shaded in including minus 11 but not including minus 4 so if you could get this problem you I think you can handle pretty much any rational any rational equation problem hope you enjoyed that