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. -4/(1-x)<5

Accepted Answer

Welcome back. I am so glad you're here. We're asked to solve the following inequality and to express the answer in internal notation, our inequality is negative 18 divided by the quantity of three minus X is all less than four. Our answer choices are answer choice. A open parentheses, negative infinity to three closed parentheses in union with open parentheses, 15 halves to infinity, closed parentheses. Answer choice B open parentheses, negative infinity to negative three closed parentheses in union with open parentheses, 15 halves to infinity, closed parentheses. Answer choice C open parentheses, 3 to 15 halves, closed parentheses and answer choice. D open parentheses, negative 3 to 15 halves closed parentheses. All right. So to solve for an inequality, we need to do two things. First, we need one side to be equal to zero and we need the other side to just have a fraction. And so here we can start by subtracting four from both sides. So then the right hand side will be equal to zero. Great. But now on the left, we have negative 18 divided by three minus X minus four, which is all less than zero. But we don't just have one fraction on the left hand side. So we need to do the subtraction indicated on the left when we are subtracting and we have a fraction, we need to have a common denominator. The common denominator here is three minus X. So we'll multiply the four by three minus X divided by three minus X. And we will have negative 18 divided by three minus X minus in the numerator, four multiplied by the quantity of three minus X divided by three minus X all still less than zero. And now we have a common denominator. So we can subtract the numerators. We'll have negative 18 minus four multiplied by the quantity of three minus X all divided by three minus X all still less than zero. Now, we can distribute this negative four to the three and the negative X in the parentheses. So we'll still have negative 18 and then negative four, multiplied by three is negative 12 and negative four multiplied by negative X is a positive four, X all still divided by three minus X still less than zero. Now, we can combine these like terms negative 18 minus 12 in the numerator that will be a negative 30. So we'll have negative 30 plus four X divided by three minus X is less than zero. And now it's in the proper format and we can solve the inequality. So what we do is we set both the numerator and the denominator equal to zero. So we'll start with the numerator negative 30 plus four, X equals zero. We can add 30 to both sides and we'll have four, X equals a positive 30. Now, we can divide both sides by four and we have X equals 34th. Both of those are divisible by two. So dividing the numerator by 2 30 divided by two is and the denominator four divided by two is two. So one of our end points is halves. Now we'll find the other end point setting our denominator equal to zero. We have three minus X equals zero will subtract three from both sides. And then we have a negative X equals negative three. Divide both sides by negative one. And we have our other end point X equals negative three divided by negative one is a positive three. So what do these end points tell us? Well, those are the end points for our intervals. If we draw a number line from negative infinity on the left to positive infinity on the right. And then we note our end points. The first one is going to be a positive three and the second one is going to be 15 halves. We have three intervals, the first from negative infinity to positive three, the second from 3 to 15 halves and the third from 15 halves to positive infinity. We are going to choose a test point from each of those three intervals, plug that test point in to our inequality. And see if the inequality is true. If it is true, then that test point tells us that that interval is true for this inequality. If it's not true, then that interval doesn't work. So for the interval from negative infinity to three, I'm going to choose the test 30.0 for the interval from to 15 halves. I'm going to choose the test 150.4 and for the interval from halves to positive infinity, I'm going to choose the test 150.10. So now we can plug those in for X for our inequality. So we'll have a negative 30 plus four multiplied not by X but by zero, all divided by three minus not X but zero, all supposedly less than zero. Simplifying four multiplied by zero is zero. So we have negative 30 in the numerator divided by three minus zero is three and negative 30 divided by three is a negative 10 is negative 10 less than zero. Yes, it is. So this interval does work for our inequality. Now let's check the next one. Plugging in four. We have negative 30 plus four multiplied not by X but by four. And that's all divided by three minus not X but four. And that's all supposedly less than zero. Simplifying four, multiplied by four is 16. So we have negative 30 plus 16, all divided by three minus four is negative one. Simplifying further negative 30 plus 16 is a negative 14 divided by negative one, negative 14 divided by negative one is positive 14 is positive 14 less than zero. Nope, it is not. So this interval does not work. Now let's check our final interview interval with the test 0.10, negative 30 plus four multiplied not by X but by 10 all divided by three minus not X but 10 simplifying four multiplied by 10 is 40. So we have negative 30 plus 40 all divided by three minus 10 is negative seven. Simplifying negative 30 plus 40 is positive 10 divided by negative seven. Now, we can keep that as a fraction is negative 17th, less than zero. Yes, it is. That's a negative number. So this interval works. Now how do we express this in interval notation? Well, the first interval that works is from negative infinity to three. For negative infinity, we always start that with an open parentheses. Parentheses means we can't quite obtain that and we can never get as far as negative infinity. Then we have a comma and then we have three is three included while we look at our original inequality that was just less than not less than or equal to. So three is not included because three made it well, three actually came from the denominator and that made it undefined. So three is definitely not included. So it gets a parentheses to close, that's in union with. And now for our next interval from 15 halves to infinity. 15 halves is not included because 15 halves would make this equal to zero. It came from the numerator when we set that equal to zero and solved. So 15 halves gets parentheses as well. Then we have a comma to positive infinity. Close that with a parentheses. And so we look at our answer choices and this matches with answer choice. A 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. -4/(1-x)<5

Key Concepts

Rational Inequalities

Interval Notation

Testing Intervals

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