Solve each inequality in Exercises 65–70 and graph the solution s...

Question

Solve each inequality in Exercises 65–70 and graph the solution set on a real number line.

1/(x + 1) > 2/(x - 1)

Accepted Answer

Hey everyone in this problem. Using a number line where us to graph the solution set for the following inequality are given five divided by x plus five is greater than 10 divided by x minus five. Alright, so writing this out, we have five over X plus five is greater than 10 over X. Whoops. Not 10 X minus five. All right. We have an inequality like this. What we want to do is we want to get everything on one side. Okay? We want to have a zero on the other side. Okay? It's easier to compare a function to zero. Okay, because we can look at the roots of that function, kind of see where it goes positive, where it goes negative. It's easier to compare to zero than it is to compare to some other number or some other function. So let's move everything to the left for now and start there. So we have five over X plus five. Okay we're gonna subtract 10 over X -5 to move it over. Okay, it's just like rearranging equation. Okay with the inequality we rearrange just like we re arrange an equation. Okay, the inequality sign stays the same when we're adding and subtracting and now we have this on the left hand side compared to zero on the right hand side. Now we want to get this as one kind of bigger function um where we can factor and things are kind of more multiplied together rather than added and subtracted because that makes it easier to find those roots and compared to the zero. So let's find a common denominator here. If we multiply five by X -5 and 10 by X plus five then we have a common denominator of X plus five times x minus five. And again this is greater than zero. Now you see that we have this factor of five in both terms five and 10. Okay, we can divide the whole thing by five. Left hand side and right hand side. We're gonna be left with X -5 - times x plus five over X plus five times x minus five. And we have to divide the right hand side by five as well. In this case +00 divided by five is still zero. All right, let's simplify this numerator, X minus five minus two X minus 10 divided by X plus five. X minus five is greater than zero. Alright, so we're getting close you just need to simplify And we're gonna find that negative X - Over X-plus five times X -5. We want that greater than zero. No, we could leave it like this and do our analysis from this function. I'm going to get rid of that negative in the numerator. Ok, sometimes it's easier just to work with positive values. Um just a little bit makes a little bit more sense but you can leave it like this with the negatives if you want as well. Okay, so I'm going to factor out that negative and I'm going to divide by it now when I divide by a negative. Okay, the sign is going to flip. So we're going to end up with X plus Over X-plus five Times X -5 is less than zero. Okay, so we divide by a negative. That sign flips All right. You can also think of it. If you wanted to kind of add this to the other side instead of subtracting then you would add it over here in the sign would be in the opposite direction. That is more straightforward. You can think of it that way as well. Okay, so we have our function here. Okay, we c we have kind of three factors. We have X plus 15 in the numerator and then we have X plus five times X minus five in the denominator. So that gives us some points of interest. Okay, from the numerator. We have the 0.-15. Okay, from the denominator we have negative five and positive five. Ok, That's where these factors will be zero. Okay, and when these factors are zero, that's where we're going to have a changing of sign. Okay, one of these has to go through zero to go from either a positive function to a negative function or a negative function to a positive function. So those are the points that we're interested in looking at. Alright, so we have our function we have our points of interest. We're gonna go ahead and call this function here. G of X. Ok. So that we don't have to keep rewriting. Alright, so let's draw a number line and get started on that. So this great big number line here we have zero positive five. And sorry for the um kind of Uneven distribution of numbers. I just wanted to fit negative 15 in there. They should be equally spaced. Alright. And we have negative five here. Alright, so we've drawn our points of interest negative 15 negative five and positive vibe. We want to kind of look at what happens on the interval surrounding these points. Okay, so let's start over here. What happens to the left of negative 50? So choose a point to the left of negative 15, say negative 20. Okay, if we have X is negative 20 then X plus 15 is going to be negative. It's gonna be less than zero negative 20 plus 15 negative five. X plus five. Well that's also gonna be negative and x minus five is also gonna be negative. So our numerator is X plus 15. Less than zero. X plus five. Less than zero. X minus five. Less than zero. So our denominator will call it D. Okay is X plus five times X minus five. So we have a negative times a negative. So our denominator is actually going to be positive, negative times a negative is a positive. So what does that mean for G. That means that G F X. Here we have a negative divided by a positive is going to be less than zero. That's actually what we want. We want G F X less than zero. So that's good. We want to include this interval In our solution. So everything to the left of negative is going to be included in our solution. And if we want to write this in terms of an interval We have negative infinity to negative and we're using a round bracket and an open circle because we are not including negative if we include negative 15, okay then our function is on the left hand side is equal to zero but that is not less than zero. Okay, we don't have less than or equal to zero. We just have strictly less than zero. So that point is not included. Alright, so we've done our first interval. Let's move to the next Between -15 and -5. So let's choose a point. Say negative 10 X plus 15 is going to be positive. Okay, if we have negative 10 negative 10 plus 15 is positive. Five is bigger than zero. X-plus five is going to be negative And X -5 is also going to be negative. Okay, so again our denominator, if we have a negative times a negative is going to be positive. Now in this case our numerator is positive, our denominator is positive. So the function G is going to be a positive divided by a positive. It's going to be bigger than zero. Okay. We actually don't want G f X bigger than zero. Okay. We want our inequality G f X less than zero. So this is not a part of our solution. Alright, moving to the next interval between -5 and five. Okay, X plus 15. Okay, we can choose a point say zero. X plus 15 is going to be positive. X-us five is also going to be positive. X minus five is going to be negative. K zero minus five is gonna be negative five. That's going to be less than zero. So our denominator we have a positive times a negative, It's going to be negative less than zero. So if our numerator is positive, our denominator is negative. A positive divided by a negative is going to give us a negative. So G f X is going to be less than zero. This is what we want, right? That's our inequality G of X. Less than zero. So between negative five and positive five, we want to include this in our solution. Now, the end points we're going to leave as open circles. Okay? Because if we're at negative five or positive five we get a zero in the denominator or function on the left hand side doesn't exist. So we don't want to include those. And if we write this as part of our interval and we're going from negative five positive five. And again we're not including the endpoints Alright? We only have one interval left and that's for everything bigger than five. So choose any point bigger than five. You can choose 10 X-plus 15 is going to be positive, X-plus five is going to be positive and X minus five is also going to be positive so everything is positive which means that G. Of X is also going to be positive or bigger than zero, which is not what we want. We want G of X less than zero. So this does not satisfy our inequality so we don't include those points. So that's it. Okay we have our number line here, we've included everything to the left of negative 15. We've included everything between negative five and five. Okay? We've written our interval Okay, negative infinity to negative 15 And negative 5-5. Not including the end points. So if we go back up to our answer choices, we see that B is going to correspond to the answer. We found negative infinity to negative 15. That the union of negative 5 to 5 and the number line looks just like we've drawn it. Thanks everyone for watching. I hope this video helped see you in the next one

Solve each inequality in Exercises 65–70 and graph the solution set on a real number line. 1/(x + 1) > 2/(x - 1)

Key Concepts

Inequalities

Rational Expressions

Graphing Solution Sets

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