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

Question

Solve each rational inequality. Give the solution set in interval notation. 10/(x+3)≥1

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 16 divided by the quantity of X plus five is greater than, or equal to four. Our answer choices are answer choice. A open bracket, negative five to negative one closed parentheses. Answer choice B open parentheses, negative five to negative one close bracket. Answer to ac open parentheses, negative infinity to negative five close bracket in union with open parentheses, negative one to infinity, closed parentheses and answer choice D open parentheses, negative infinity to negative five close parentheses in union with open bracket, negative one to infinity, close parentheses. All right. In order to solve this inequality, we need to do two things. We want to have one side equal to zero. And then we want the other side to be just a single fraction. And once those two conditions are met, we can start to actually solve the problem. So let's get one side equal to zero and we'll take four from both sides subtracting four from both sides of the inequality. So on the left, we'll have 16 divided by X plus five minus four. And that's all still greater than, or equal to. And on the right four minus four is zero. All right, one condition met. But now on the left, we're subtracting. And when we're subtracting, and one thing is a fraction, we need to have a common denominator. The common denominator here will be X plus five. So we need to multiply the four by X plus five divided by X plus five. And so we'll still have 16, divided by X plus five. And now we're subtracting four, multiplied by X plus five in the numerator. And that's divided by X plus five in the denominator all still greater than or equal to zero. Now, we have a common denominator so we can combine our numerators, we'll have 16 minus four multiplied by the quantity of X plus five. And that's all divided by our single denominator. X plus five, all still greater than or equal to zero. Now, we can distribute this negative four to the X and the five in parentheses. So we'll have still 16 and now negative four, multiplied by X is negative four, X and negative four, multiplied by five is negative 20. All still divided by X plus five, all still greater than or equal to zero. Now, combining like terms in the N writer, we have 16 minus 20 and minus 20 is a negative four and then negative four minus four. X all still divided by X plus five, all still greater than or equal to zero. But now those two conditions are met, we have a single fraction on the left and zero on the right now, we can begin to solve the inequality. So what we do now is we're going to set both the numerator and the denominator equal to zero. Those will give us the end points of our intervals. So let's start with the numerator negative four minus four, X equals zero, adding four to both sides of the equation. On the left, we have negative four X that's equal to zero plus four is positive four. On the right now, dividing both sides by negative four on the left, we'll have X Y and on the right four divided by negative four is a negative one. So one of our endpoints is at negative one. Let's find the other setting our denominator equal to zero X plus five equals zero. Subtracting five from both sides of our equation. On the left, we have X ya on the right, zero minus five is negative five. Our other end point is negative five. Now we'll place these two end points on a number line from negative infinity on the left to positive infinity on the right. The first end point we encounter from the left from negative infinity is going to be the negative five and the next one will be the negative one and then up to positive infinity. Now we have three intervals, the first one from negative infinity to negative five, the next interval from negative five to negative one. And the final interval from negative one to positive infinity, we will choose a test point from each interval. Plug that test point value in for X in our inequality. If the inequality is true for the test point, then that interval is included. If the inequality is not true, then that test point tells us that that interval is not included. So for the interval from negative infinity to negative five, I'm going to choose the test point negative six. For the interval from negative five to negative one. I'm going to choose the test point negative two. And for the interval from negative one to positive infinity, we're going to choose the test 10.0. So now let's plug those test points in for X beginning with the test point negative six, we'll have negative four minus four not multiplied by X but now multiplied by negative six all divided by not X but now negative six plus five, supposedly I greater than or equal to zero. Now simplifying in the numerator, negative four, multiplied by negative six is a positive 24. So we have negative four plus 24 in the numerator and in the denominator negative six plus five is negative one. Simplifying negative four plus 24 is a positive 20 divided by negative 1 divided by negative one is negative 20 is negative 20 greater than or equal to zero. It is not. So because that is not true, our test point tells us that this interval is not included. Now, let's test our next test point negative two. For the interval from negative five to negative one, we'll have negative four minus four, not multiplied by X but multiplied by negative two, all divided by not X but negative two plus five, all supposedly greater than are equal to zero. Simplifying in the numerator negative four multiplied by negative two is a positive eight. So we have negative four plus eight in the denominator, negative two plus five is a positive three. Simplifying negative four plus eight. In the numerator is a positive four. That's divided by three. We can keep that as a fraction is four thirds greater than or equal to zero. Yes, it is. So this test point works and tells us that interval is true for this inequality. Now let's test the last interval. So we'll have negative four minus four, not multiplied by X but multiplied by zero, all divided by not X but zero plus five, all supposedly greater than or equal to zero. Now, simplifying in the numerator, negative four, multiplied by zero is zero. So we just have negative four in the numerator as divided by, in the denominator zero plus five is positive five. We can keep that fraction as negative 4/5. Is it true that negative 4/5 is greater than or equal to zero. It is not. So that tells us that this interval is not included. So we know that it is the interval from negative five to negative one where this inequality is true. Now we have to express this in interval notation. So we want to take a look at our original inequality and we notice that it is greater than or equal to zero. So any values that make it equal to zero are included as well. Now our end points, where did those come from? Well, the negative five came from setting the denominator equal to zero. And if our denominator is equal to zero, then the whole thing is undefined and we cannot have that. So even though it's greater than or equal to zero, that negative five cannot be included. Now to express this with interval notation, we say that something is not included by using a parentheses. So we'll put open parentheses and the negative five and then comma to say it goes all the way up to our next endpoint of our interval which is negative one is negative one included. Well, we got negative one when we set the numerator equal to zero. And as we said, it can be equal to zero, it's greater than or equal to. So the negative one can be included. We express that with a bracket bracket says you can be included. So we close that negative one with a closed bracket and that's how we express this inequality, the solution for it. 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. 10/(x+3)≥1

Key Concepts

Rational Inequalities

Domain Restrictions

Interval Notation and Sign Analysis

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