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. 10/(3+2x)≤5

Accepted Answer

Welcome back. I am so glad you're here. We're asked to solve the following inequality and express the answer in interval notation. Our inequality is 15 divided by the quantity of one plus four X is less than or equal to six. Our answer choices are answer choice. A open parentheses, negative infinity to negative 1/4 close bracket. In union width, open parentheses 3/8 to infinity, closed parentheses. Answer choice B open parentheses, negative infinity to negative 1/4 close parentheses in union width, open bracket, 3/8 to infinity, closed parentheses, answer choice C open parentheses, negative infinity to negative 3/8 close bracket in union width, open parentheses, 1/4 to infinity. Close parentheses. In answer choice. D open parentheses, negative 1/4 to 3/8 close bracket. All right. When we have fractions as part of our inequality, we need to do a couple of things before we can begin to solve for it. We want to set one side equal to zero and then we want to ensure that we have one fraction on the side that's not zero. So we'll start off, we'll subtract six from both sides so that the right hand side will be zero. And then on the left, we have 15 divided by one plus four, X minus six is less than, or equal to six minus six is zero. So we've got one side equal to zero. But now on the left, we need just one fraction. In order to subtract. When we have a fraction, we need to have a common denominator. The common denominator will be one plus four X. So we multiply the six by one plus four X divided by one plus four X. So we will have 15, divided by one plus four X still minus in the numerator six, multiplied by the quantity of one plus four X all divided by one plus four X is still less than or equal to zero. And now that we have a common denominator, we can subtract our numerators. The numerator will be 15 minus six multiplied by the quantity of one plus four X all over the single denominator of one plus four X still less than are equal to zero. Now, we can distribute that negative six to the one and the four X in the parentheses. So we'll still have that 15 and now negative six, multiplied by one is negative six and negative six multiplied by four, X is negative 24 X still over one plus four X in the denominator and all still less than are equal to zero. Now, we can combine the light terms in the numerator 15 minus six will give us a positive nine. So we have nine minus 24 X divided by one plus four. X is less than or equal to zero. And now the conditions have been met for us to solve for the following inequality. All right. So in order to do that, we are going to set both our numerator and our denominator equal to zero. That will give us the end points of our intervals. So let's start with the numerator nine minus 24. X is equal to zero. Now we will solve for X going to subtract nine from both sides. And on the left, we have negative 24 X that is equal to zero minus nine is negative nine. Now I'll divide both sides by negative 24. And on the left, we'll have X equals Y solving for X and on the right, negative nine divided by negative 24 will be a positive nine 24th and both the numerator and the denominator are divisible by three. So this simplifies two in the numerator nine, divided by three is a positive three. And in the denominator 24 divided by three is a positive beat. So one of our end points is X equals 3/8. Now let's find the other end point. We're setting our denominator equal to 01 plus four, X equals zero. Subtracting one from both sides. On the left, we have four X equals and then on the right zero minus one is negative one. We divide both sides by four. On the left X is by itself, we have X equals ya. And on the right, we have negative 1/4 there's our other end point for our intervals. So now let's set up a number line going from negative infinity on the left to positive infinity on the right. And the end points for our intervals be besides negative infinity and positive infinity are from the left. The first one that we encounter is negative 1/ and the next one we encounter is a positive 3/8. So we have three intervals, one from negative infinity to negative 1/4 1 from negative 1/4 to positive 3/8 and one from positive 3/ to positive infinity. We are going to choose a test point from each of those three intervals. We're going to plug that test point in for X in our inequality. And we're going to see if the statement is true. If it's true, then that interval is included. If it's not true, then that interval is not included. So for the test points for, for negative infinity to negative 1/4 I'm going to choose negative one for the test point for the interval from negative 1/4 to positive 3/8. I'm going to choose zero. And for the test point from positive 3/8 to positive infinity, I am going to choose a positive one. So now we'll begin to plug those in for our inequality. So we have for the test point negative one, we'll have nine minus 24 not by multiplied by X. But now multiplied by negative one in our numerator that's divided by, in the denominator one plus four, not four X, but now four multiplied by negative one. And this is all supposed to be less than or equal to zero. Now simplifying negative 24 multiplied by negative one is a positive 24. So we have nine plus 24 in the numerator divided by, in the denominator four, multiplied by negative one is negative four. So we have one minus four. Simplifying further nine plus 24 in the numerator is a positive 33 that's divided by one minus four in the denominator which is a negative three and divided by negative three equals a negative 11. And our negative 11 is supposed to be less than or equal to zero. Is that true? Yes, it is. So this interval is included for our inequality that one works. Now we can test the next interval. We'll use our test point of zero. So we have nine minus 24 not multiplied by X but multiplied by zero, divided by one plus four, not multiplied by X, but now multiplied by zero. And this is all possibly less than or equal to zero. Simplifying negative 24 multiplied by zero is zero. So we just have nine in the numerator and four multiplied by zero is zero. So we just have one in the denominator nine divided by one is nine, is nine less than, or equal to zero. Nope, it's not. So this interval does not work. It is not included. Now, let's test our last interval with the test 00.1, we have nine minus 24 multiplied not by X but by one, that's all divided by one plus four, multiplied not by X but by one. And we'll see if that's all less than, or equal to zero in the numerator negative 24 multiplied by one is negative 24. So we have nine minus 24 in the denominator four multiplied by one is four. So we have one plus four, nine minus 24 in the numerator is a negative 15. That's divided by one plus four in the denominator which is five, a negative 15 divided by five is a negative three is negative three less than or equal to zero. Yes, it is. So this interval works, it is included for inequality. So now we have to write those two intervals that are included in interval notation. So for the first one, we're talking about from negative infinity all the way up to a negative 1/4. Well, for Interpol notation, we're using either parentheses or brackets, parentheses. Say that something is not included, brackets, say that it is included. And it's important to note that in our original inequality, we're looking for something that is less than or equal to. So if that test point made it equal to zero, then it is included. So how about from negative infinity? Well, negative infinity is not included. We can never quite get so low as negative infinity. So we put a parenthesis for negative infinity. Then a comma to say going all the way up to negative 1/4 and now is negative 1/4 included. We know that we set that um we found negative 1/4 when we set the denominator equal to zero. But if our denominator equals zero, then the whole fraction is undefined. And we cannot have that. So negative 1/4 cannot be included. We have to close that with a parentheses. Now we say that is in union with because it also includes the next interval which is from positive three eights all the way up to positive infinity. And from 3/8 was found when we set our numerator equal to zero. And because this inequality is less than or equal to zero, and if the numerator equals zero, then the whole fraction equals zero. That's all cool. So 3/8 can be included. So 3/8 gets a bracket, bracket, 3/8 comma all the way up to infinity and infinity can never be obtained. So that's closed with the parentheses. And that is our inequality expressed in interval notation. We look at 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. 10/(3+2x)≤5

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation

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