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

Question

Solve each rational inequality. Give the solution set in interval notation. See Examples 8 and 9. 3/(x-2)<1

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 three is less than five. Our answer choices are answer choice. A open parentheses, negative infinity to negative eight closed parentheses in union with open parentheses. Three to infinity, closed parentheses. Answer choice B open parentheses, negative infinity to negative eight closed parentheses in union width, open parentheses, negative three to infinity. Closed parentheses. Answer choice C open parentheses, 3 to 8, eight closed parentheses and answer choice D open parentheses, negative infinity to three closed parentheses in union with open parentheses negative or excuse me. Eight to infinity, closed parentheses. All right. To solve our inequality, we need two conditions met first, we need to have zero on one side of the inequality and the other side needs to be one fraction. So to get zero on one side, we can subtract five from both sides of the inequality. And then on the left, we'll have still the fraction 25 divided by X minus three and that's minus five. And that's less than now, on the right hand side, five minus five is zero. The first condition is met, but we don't have one single fraction on the left anymore. Now, we need to do the subtraction. And when we're subtracting, subtracting with fractions, we need to have a common denominator, the common denominator will be X minus three. So we'll multiply this five by X minus three divided by X minus three. So we will have 25 divided by X minus three minus in the numerator five multiplied by the quantity of X minus three divided by X minus three. All still less than zero. Now, we have a common denominator and we can subtract our numerators. We will have 25 minus five multiplied by the quantity of X minus three, all divided by X minus three, all still less than zero. Now, we can just simplify the numerator. Let's distribute this negative five to the X and the negative three in parentheses. So, rewriting our 25 now multiplying negative five, multiplied by X is negative five, X and negative five, multiplied by negative three is a positive 15. That's all divided by X minus three and all less than zero. We're close. Now, let's combine the like terms in the numerator 25 plus 15. 25 plus 15 is 40. So we have 40 minus five X all divided by X minus three, all less than zero. And all right, we've met those two conditions that we mentioned a while ago. Now we can solve for the inequality. What we need to do is set both the numerator and the denominator equal to zero. Those will give us the end points of our intervals. So let's do that. Setting our numerator equal to 0 40 minus five X equals zero. We can subtract 40 from both sides and we'll have negative five X equals negative 40. We can divide both sides by negative five and we'll have X equals on the left and on the right, negative 40 divided by negative five is a positive 81 of our end points is eight. All right. Next, we'll set the denominator equal to zero, X minus three equals zero. We can add three to both sides and we'll have X equals a positive three. There's our other end point. What do we do with these end points? We can draw a number line from negative infinity on the left to positive infinity on the right. We can note our end points. So the first one coming from negative infinity is going to be the positive three. And then the next endpoint is positive eight. And we now have three intervals, one from negative infinity to 31 from 3 to 8 and one from eight to positive infinity. We are going to choose test points from each one of those intervals and we are going to plug the test point in for X in our inequality. If the inequality is true, the test point tells us that that interval is true for the inequality. So for the interval from negative infinity to three, I'm going to choose the test 30.0 for the interval from 3 to 8. I'm going to choose the test 80.4 and for the interval from eight to infinity, I'm I am choosing the test 80.10 and now we'll plug those in for X in our inequality, we'll have 40 minus five multiplied not by X but now by zero, that's all divided by not X but zero minus three. And that's all supposedly less than zero. Let's simplify negative five, multiplied by zero is zero. So we just have 40 in the numerator divided by zero minus three in the denominator is negative three. We can keep it as a negative 43rd because we just need to know is that less than zero. And it is. So the interval from negative infinity to positive three works for this inequality. Now let's check the next interval with our test 30.4, 40 minus five multiplied not by X but by four, all divided by not X but four minus three, supposedly less than zero, negative five multiplied by four is negative 20. So we have 40 minus 20 in the numerator divided by four minus three in the denominator is a positive one. Simplifying 40 minus 20 is a positive 20. So we have 20 divided by one which is 20 is 20 less than zero. No, it is not. So this interval is not true for our inequality. Let's test the final one. Plugging in the test 10.10 40 minus five multiplied not by X but by 10 all divided by not X but 10 minus three, all supposedly less than zero. Simplifying negative five multiplied by 10 is negative 50. So we have 40 minus 50 in the numerator all divided by 10 minus three in the denominator is seven, 40 minus 50 is a negative 10. So we have negative 17th. Is that less than zero? Yes, it is. So this inequality is also true. Now how do we say this in interval notation? We want the interval from negative infinity to three and from eight to positive infinity. Well, we are always going to use an open parentheses for negative infinity. A parentheses tells us that we can never quite obtain that number and we can't quite get to negative infinity. Then we have a comma and we have three. Now, what do we close three with is three included or not in this inequality? Well, we go back to what we had and this is just less than unless they and doesn't include things that are equal to zero. And notably, we got that three when we set the denominator equal to zero. So if we included three, then this would be undefined and we cannot have that. So three is definitely not included. Three is a parentheses because there are two unconnected intervals we say that's in union with and now for the eight, does the eight get a parentheses? Well, yes, it does because it just needs to be less than zero. It cannot be equal to zero. We got the eight when we set the numerator equal to zero and we can't have it be equal to zero. So eight also gets a parentheses and then we have eight comma infinity. Infinity. Also gets a parentheses because we can never quite obtain infinity. All right, we look at our answer choices and this matches with answer choice. D 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-2)<1

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation

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