Solve each inequality. Give the solution set using interval no...

Question

Solve each inequality. Give the solution set using interval notation. $\frac{5 x + 2}{x} + 1 < 0$

Accepted Answer

Hey, everyone in this problem, we're asked to find the solution set of the following inequality and express it in interval notation. The inequality we're given is 14 X plus 20 divided by X plus two is less than zero. We're giving four answer choices A through D and they're all just different intervals for the solution set. And we're gonna come back to those as we work through this problem. Now writing what we were given, we have 14 X plus 20 all divided by X, right. And then we have this two term at the end and this is all less than zero. Now, in order to find the solutions, what we wanna do is get this left hand side to be one rational function. OK? We want everything to be multiplied and divided and we want all of these factors multiplied and divided. We don't want this plus two term at the end here. OK. So in order to combine this plus two term, the rest of our function, we need to find a common denominator. The denominator for our first term is X. So we want this to, to have a denominator of X as well. In order to do that, we need to multiply the denominator by X. If we multiply the denominator by X, we have to also multiply the numerator by X. So what we get is 14 X plus plus two X all divided by X. OK. And this is less than zero. So now you are our common denominator. We've simplified this into one rational function. And now we just want to simplify that numerator and we have 14 X plus two X. So that can be written as 16 X and then we have our plus 20 as well. And all of this is divided by X. There's been less than zero. Now, we can simplify one step further here. This is not necessary, but it makes it a little bit simpler. We have 16 and 20. These two terms in the numerator, both of them are divisible by four. So we can divide the entire equation left-hand side and right hand side by four. OK. We get four X plus five divided by X on the left hand side and on the right hand side, we have zero, we divide it by four and we just get zero. All right. Now, there's a couple of different approaches we could take with this problem at this stage. OK? We can multiply both sides by X here. And then we'd have four, X plus five is less than zero. We have to be a little bit careful there. OK? We need to take into consideration the restriction that X cannot be equal to zero because we have an X in the denominator, we don't want to divide by zero. We also have to take into the take into account what happens if X is negative, if X is negative or multiplying by a negative value and an inequality which is gonna flip our inequality. OK. So you're gonna have to consider two cases if you take that approach instead of taking that approach and having to worry about those different cases and the restrictions, what we're gonna do is look at our points of interest and our points of interest are gonna be any points where either the numerator or denominator of the left hand side is equal to zero. So if we look at the numerator, we have four X plus five. If we set that equal to zero and solve for X, we get that X is equal to negative five quarters. OK? So that's gonna be one point of interest and the other is gonna be from setting the denominator equal to zero. That's just gonna give us that X is equal to zero. So those are our two points of interest. And what we're gonna do is add those to a number line and we're gonna evaluate what happens on either side of these points of interest. OK? So we have negative five quarters, we have zero and then we have everything on either side. Now, if we give XX less than negative five quarters, any of that means X is more negative. So you could imagine substituting at a point like X equals negative two, well, four X minus five in this case or sorry, four X plus five is gonna be less than zero. It's going to be negative in X, the X value is also negative or less than zero. So what that tells us is that the left hand side of our equation. OK. I'm gonna call this L H S. And when I say L H si mean this four X plus five divided by X, we have a negative divided by a negative. So it's gonna be positive. Now, our inequality wants that left hand side to be less than zero. OK? On this interval, it's greater than zero. So we're gonna draw a red X on this interval. This is not gonna be part of our solution set. Now we move to the next interval. OK? And it's gonna be between negative five quarters and zero. So you can imagine choosing a point, say like negative one. If you substitute that into four X plus five, it's gonna give you a positive value. Your four X plus five is greater than zero, but X is still gonna be negative. OK, X is less than zero. So we have a positive value divided by a negative value. The left hand side is going to be negative and that's exactly what we want for our inequality. So this interval is gonna satisfy the inequality and it's gonna be part of our solution set. We're gonna put a green check mark there. Now we also wanna evaluate these end points. OK? At negative five quarters. Do we include that point or do we not include it? Well, if X is equal to negative five divided by four and the left hand side is gonna be equal to zero, but zero is not less than zero. OK? We have a strict inequality here. We don't have less than, or equal to. So we don't want to include that point. I'm just gonna draw a open circle on our number line to indicate that we're not including that point. And that's just gonna be a reminder to us when we go to write this in interval notation what to do at that point. Now, when X equals zero, OK. Well, that's a restriction. Remember if X is equal to zero, we're dividing by zero, the function is undefined. So that's also not going to be part of our solution set. So we draw an open circle there as well. All right. Now we move to our last interval. And that's gonna be the case where X is greater than zero. If X is greater than zero, then four X plus five will be greater than zero, X will also be greater than zero. And so the left hand side, a positive value divided by a positive value will also be positive or greater than zero. And again, that's not gonna satisfy our inequality. We want it to be less than zero. And so this is not part of our solution set, we can cross it out with a red X. So now we have our solution, we just need to convert to interval notation. So we found that there was only one interval for this solution set and one interval where the inequality is satisfied and that interval is from negative five quarters up to zero. And we're using round brackets on both of the end points to indicate that we do not include them. If we go up to our answer choices and compare what we've just found right off the bat, we can eliminate option B and C because they both have the union of two intervals. And we found that the, we only have one interval in our solution. OK? And the correct answer is gonna be option A. OK. This interval matches what we found. It's the interval from negative five quarters to zero. Thanks everyone for watching. I hope this video helped see you in the next one.

Solve each inequality. Give the solution set using interval notation. $\frac{5 x + 2}{x} + 1 < 0$

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation

Watch next

Solve each inequality. Give the solution set using interval notation. 5x+2x+1<0\frac{5x + 2}{x} + 1 < 0

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation

Watch next

Solve each inequality. Give the solution set using interval notation. $\frac{5 x + 2}{x} + 1 < 0$