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

Question

Solve each rational inequality. Give the solution set in interval notation. See Examples 8 and 9. 3/(x-6)≤2

Accepted Answer

Welcome back. I am so glad you're here. We are asked to solve the following inequality and express the answer in interval notation. Our inequality is the fraction divided by the quantity of X minus eight which is less than or equal to five. Our answer choices are answer choice. A open parentheses, negative infinity to eight close bracket in union with open parentheses, 12 to infinity, closed parentheses. Answer choice B open parentheses, negative infinity to eight closed parentheses in union with open bracket. 12 to infinity, closed parentheses. Answer choice C open bracket, 8 to 12 closed parentheses and answer choice D open parentheses, 8 to 12 closed bracket. All right. Before we can start to solve for this inequality, we need two conditions to be met. One is that there needs to be zero on one side of the inequality. And the other is that there needs to be just one single fraction on the other side, we can start by getting zero on one side by subtracting five from both sides of the inequality on the left. We'll still have the fraction 20 divided by X minus eight and that is minus five. And that's all still less than or equal to. And now on the right five minus five is zero. Well, we've got one condition. Matt, we have the zero, but we don't have one single, single fraction on the left anymore. We need to do the subtraction. And when we're subtracting with fractions, we need to have a common denominator, the common denominator will be X minus eight. So we need to multiply this five by X minus eight divided by X minus eight. So now we'll have 20 divided by X minus eight minus five, multiplied by the quantity of X minus eight, all divided by X minus eight, all still less than are equal to zero. Now, we have a common denominator and we can subtract across our numerators. We will have 20 minus five multiplied by the quantity of X minus eight all divided by X minus eight, all still less than or equal to zero. Now, we can simplify in the numerator. We can take negative five and distribute it to the X and the negative eight in the parentheses. We can bring over the 20 and now negative five, multiplied by X is negative five X and negative five multiplied by negative eight is a positive 40. That's all divided by X minus eight, all less than or equal to zero. Still now combining the like terms in the numerator plus 40 20 plus 40 is 60. So we have 60 minus five X in the numerator divided by X minus eight in the denominator all still less than or equal to zero. And now we have met those conditions that we talked about a while ago where we have a single fraction on one side and zero on the other. Great. OK. Now, to solve for the inequality, what we do is we set both the numerator and the denominator equal to zero. And we solve for X and those are going to be the end points of our intervals. So let's do that 60 minus five X, our numerator setting that equal to zero, we can subtract 60 from both sides and we'll have negative five, X equals negative 60. Divide both sides by negative five and we have X equals a positive 12. There's one of our end points, let's find the other. We'll set our denominator X minus eight equal to zero. We'll add eight to both sides and our other end point is X equals eight. Now, what do we do with these end points? Well, we can set up a number line from negative infinity on the left to positive infinity on the right. And then we can note our end points. The first one from negative infinity on the left. The first one we'll run into is a positive eight. The next end point is the positive 12. And now we have three intervals, one from negative infinity to eight, one from 8 to 12 and one from 12. To positive infinity. Now, what we're going to do is choose a test point from each one of those intervals, plug the test point in for X in our inequality and see if the inequality is true. If it is true, then that interval is true for our inequality. So for the interval from negative infinity to eight, I'm going to choose the test 80.0 for the interval from 8 to 12. I'm going to choose the test 120.10 and for the interval from to infinity, I'm going to choose the test 120.20. And now plugging in those test points, we'll start with the interval from negative infinity to eight. Let's plug zero in for X. So we'll have 60 minus five, not multiplied by X but multiplied by zero all divided by not X but zero minus eight, all supposedly less than or equal to zero. Simplifying negative five, multiplied by zero is zero. So we just have 60 in the numerator divided by zero minus eight is negative eight. We can just leave it like this without simplifying too much. We have negative 68. Is that less than or equal to zero? Yes, it is. That's a negative number. So this interval is true from negative infinity to eight for our inequality. Now let's test 10. So we'll have 60 minus five multiplied not by X but by 10, all divided by not X but minus eight, all supposedly less than are equal to zero. Simplifying negative five multiplied by 10 is negative 50. So we have 60 minus 50 in the numerator and in the denominator 10 minus eight is a positive two. Simplifying 60 minus 50 is a positive 10. So we have 10 divided by two which is five is five less than or equal to zero. No, it is not. So this interval is not true for our inequality. Now let's test the final interval with the test 0.20 we have minus five multiplied not by X but by all divided by not X but minus eight, all supposedly less than are equal to zero. Simplifying negative five multiplied by 20 is negative 100. So we have 60 minus 100 in the numerator divided by 20 minus eight is a positive 12. Simplifying 60 minus 100 is negative 40. So have negative 40 divided by 12. We can leave that like that is negative 52nd less than or equal to zero. Yes, it is. That's a negative number. So this interval is true for inequality as well. All right, we have from negative infinity to positive eight and from positive 12 to positive infinity. How do we express that in interval notation? Well, we're always going to have parentheses for our infinities. So we start with open parentheses, negative infinity comma and then that goes all the way up to eight. But does it include eight? Well, we need to look back at our original inequality to see if eight is included, we have less than or equal to eight could be included. But where did it come from? Well, eight was when we set the denominator equal to zero. And if the denominator is equal to zero, this whole thing is undefined. Eight, you cannot be included. Sorry. Eight, we say that eight is not included by closing it with a parentheses. And then because we have two unconnected intervals, we say that's in union with and then we have the 12 is the included. Where did the 12 come from? It came when we set our numerator equal to zero. And so plugging 12 in, we'll set it equal to zero. And because this is less than, or equal to 0, 12 can be included. We say that 12 is included by using an open bracket. And then we say 12 comma infinity, closing infinity with parentheses because we can never quite obtain infinity. We look at all of our answer choices and this matches with answer choice. B well done. We'll catch you on the next one.

Solve each rational inequality. Give the solution set in interval notation. See Examples 8 and 9. 3/(x-6)≤2

Key Concepts

Rational Inequalities

Interval Notation

Critical Points

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