Solve each inequality. Give the solution set in interval notation...

Question

Solve each inequality. Give the solution set in interval notation. 4/(x+6)>2/(x-1)

Accepted Answer

Hey, everyone in this problem are asked to provide the solution set for the given rational inequality and interval notation, the inequality we're given is four divided by X plus 23 is greater than three divided by X minus two. We're given four answer choices A through D and each of them have the solution as a union of two sets and they just differ in what those sets are. So we're gonna come back to those answer choices as we work through this problem. But let's get started. We have four divided by X plus 23 is greater than three divided by X minus two. When we have an inequality like this, what we wanna do is get everything to one side and just have zero left on the other side. And it's a lot easier to compare a function to zero because you're looking at just whether that function is positive or negative than it is comparing one function to another function or a function to a different number. We're gonna move first, our three divided by X minus two to the left hand side. And when we do that, we have that four divided by X plus minus three divided by X minus two is greater than zero. Now, what we wanna do here too is write some restrictions. So in our denominator, we have X plus 23 and we have X minus two. So if X is equal to negative or X is equal to two, the denominator will be equal to zero. OK? So the left hand side will be undefined. We can't have a solution. And so we write these restrictions that X cannot be negative 23 and it cannot be equal to two. OK. This is a good habit just to write those restrictions right off the bat because as you simplify, you might have terms canceling out the denominator um might simplify and you won't have those original restrictions. OK? So it's good to just write them out. And then when you find solutions at the end, just double check that these values are not there. OK? All right. So simplifying on the left hand side, we want to combine this into one rational function instead of these two um terms that we have here. So what we need is a common denominator and it's just like finding a common denominator when we have just numbers. OK. Our left term has a denominator of X plus 23. The right term has a denominator of X minus two. So for our left function, in order to have a common denominator, we're gonna multiply by X minus two. OK. If we do it in the numerator, then we have to do it in the denominator. So we have four multiplied by X minus two, divided by X plus 23 multiplied by X minus two minus three. And for this function, we need to multiply by X plus 23 in the numerator and denominator. So that we have that common denominator. So X multiplied or sorry three multiplied by X plus divided by X minus two, multiplied by X plus 23. And all of this is greater than zero. Now, you can see that we have this common denominator. So each of our fractions on the left hand side has the denominator X plus multiplied by X minus two. So we can combine these and simplify the numerator. Let's start by expanding from the first term. We have four X minus eight and we're gonna subtract everything from this second term. So we have minus three, X minus 69. OK. Just expanding all of these brackets and all of this is divided by that common denominator X minus two, multiplied by X plus 23. And again, this is still greater than zero. Now, we can simplify a little bit further. OK? We have some common terms or some similar terms in the numerator. So grouping like terms, we have four X minus three X, it's gonna leave us with X and then we have minus eight minus 69 that's gonna give us minus and the denominator stays the same X minus two multiplied by X plus 23. And this is greater than zero. So now we've simplified this rational function as much as we can and we have this single rational function where all of our factors are either either being multiplied or divided. Now, we want to think about where we might get points of interest and now our points of interest, what we really wanna know is where could our function on the left hand side switch from being positive to being negative because that's where we're gonna get changes in our solution. And we're looking at whether this is greater than zero. So we want to know when it's positive. So let's find the points where it might switch from being positive to negative and that's gonna be where any of our factors equal zero. So in the numerator, if X is equal to 77 and our function is gonna be equal to zero and then on either side of 77 something different might happen. And then we have those restrictions from the denominator. OK? If either of our factors in the denominator are zero, we're gonna get the sign changing as well. And so we have X equals 77 X equals two and X equals negative 23 as our points of interest. OK? Now I've written X equals that does not mean that these are the solutions, these are just again, our points of interest. So let's write that so that we do not get confused. Let's give ourselves some more room. And now that we have our points of interest, we are gonna draw a number line and this I find helps keep our intervals separated. And it's just a good way to visualize what's going on. So we have our number line and we wanna add our three points of interest. So we have negative 23 on the left moving to the right, we get two and then further right, we get 77. OK? And this number line is not drawn to scale, but that's OK. Now, for each of these intervals, we wanna look at what happens to our function. So starting with the interval on the left, OK, to the left of negative 23 all of our X values are gonna be less than negative 23. If our X value is less than negative 23 then X minus 77 is gonna be less than zero and it's gonna be negative X minus two is also gonna be less than zero and X plus 23. Well, that's also gonna be less than zero. It's gonna be negative. So what we have is that our numerator, which we're gonna indicate with N K is negative, it's less than zero because that's just that X minus 77 term, our denominator is X minus two, multiplied by XT plus 23. OK? Both of those factors are negative, negative multiplied by negative, gives us a positive. So our numerator is negative, our denominator is positive. And so the entire left hand side, we're gonna write as LHs is going to be less than zero. OK? It's going to be negative. And that is not a solution. We want this function greater than zero. So we're gonna give this interval a red X OK. This is not part of our solution set. OK. Moving to our next interval, we're gonna do the same thing. We have X between negative 23 and positive two, X minus 77. That's still gonna be negative X minus two. It's still going to be negative but X plus is going to be positive. So we have our numerator is negative. Our denominator is also gonna be negative because we have a negative multiplied by a positive. And then we have a negative divided by a negative which is gonna give us at this left-hand side of our inequality is greater than zero. And that's exactly what we want. And so this interval satisfies our inequality. It gets a green check mark that is gonna be part of our solution set. Now, in addition to figuring out that this interval is in our solution set, we need to look at the end points, OK? Our end points are negative 23 and two. We know that we want to exclude those because those are restrictions. If X is equal to negative 23 or X is equal to two, the denominator is equal to zero and the left hand side is undefined. OK. So we're gonna draw round circles, open circles at negative 23 and two to indicate that we are not including those end points. OK? And we're gonna translate that into our interval notation when we come back to write out our full interval notation. But for now, it's just a reminder that we're not including those points. Moving to our next interval. We have X between two and 77 X minus 77 is still going to be negative X minus two is going to be positive X plus 23 is going to be positive. So again, our numerator is less than zero, it's negative. Our denominator is greater than zero, positive, a negative divided by a positive, gives us that the left hand side of our inequality is less than zero. And just like the first interval we looked at that does not satisfy our inequality. This is not a solution. So we draw a red X on that interval moving to our final interval, we have X greater than 77. And all of these factors are gonna be positive. In that case, X minus 77 is greater than zero, X minus two is greater than zero and X plus 23 is greater than zero. So for multiplying and dividing positive numbers, we are gonna get a positive result. The left hand side is going to be greater than zero, which again is what we want. So this gets a green check mark for that interval. We're left with one last thing to figure out is it 77. When X is equal to 77 do we get a solution or no? Well, if X is equal to 77 the left hand side is gonna be equal to zero. But our inequality is strictly greater than zero. It's not greater than, or equal to. So we actually don't want to include X equals 77. We draw an open circle to indicate that we exclude it because we don't want equal to zero. So now we have our entire number line drawn out and we just need to write it in interval notation. Let's give ourselves a little bit more room to convert this to interval notation. So the first interval we have is from negative 23 up to two, the end points are not included. And so we're gonna use round brackets. So we have negative 23 comma two. We're gonna take the union of that interval with the second interval we found OK. This starts at 77 and goes on up to infinity. OK? Any value to the right of 77. So we have 77 up to infinity. And again, we're using round brackets because we're not including the end points. And so this is gonna be the solution set for this problem. The union of the interval from negative 23 to 2 with the interval from 77 to infinity. Now, if we compare this to our answer choices, option C and D have the first interval starting at positive 23 instead of negative 23. And so we can eliminate those two. OK. The first interval doesn't match for option A and B. The first interval matches from negative 23 to 2. But for option A, the second interval starts at negative 77 instead of positive 77. So A is not the correct answer either. And we are left with B A which matches that solution set that we found. Thanks everyone for watching. I hope this video helped see you in the next one.

Solve each inequality. Give the solution set in interval notation. 4/(x+6)>2/(x-1)

Key Concepts

Inequalities

Interval Notation

Critical Points

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