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.
1/(x+2)≥3

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. The inequality we're given is 11 divided by X plus five is greater than or equal to two. And the answer choices we have are a from negative infinity to negative five in union with one half to infinity. And that one half has a bracket and the negative five has a parentheses be from negative infinity to negative five. This time negative five has a bracket and union with parentheses one half to infinity. Answer choice C bracket, negative 5 to 1 half parentheses and answer choice D parentheses negative 5 to 1 half bracket. Alright. Well, we recall from previous lessons that the first step in solving a rational inequality is that we want one side equal to zero. So for this one, I am going to subtract two from both sides of the inequality. So we'll have on the left 11 over X plus five minus two is greater than or equal to, on the right hand side two minus two is zero. No, still part of this first step. We need to ensure that the side that has the fractions is just one single fraction. So we have 11 over X plus five minus two, we have to do that subtraction first. And in order to subtract when we have a fraction, we need to have a common denominator. And in this case, we do not have a common denominator. We're gonna scoot this to over a little bit. What would our common denominator be? It would be this X plus five. So we're going to multiply two times the fraction X plus five over X plus five because that's one, it simplifies to one. So now we can do that multiplication. I'll bring down 11 over X plus five minus. Now, we have two times in the numerator X plus five and that's all over X plus five, all still greater than or equal to zero. We're still working on combining this to one fraction. Now, I am going to distribute this to in the second fractions numerator. So bringing down 11 over X plus five again minus. Now, simplifying the numerator two times X is two X and two times five is plus 10 still over X plus five and still greater than or equal to zero. But now we've got a common denominator and we can combine our enumerators. And so I am going to distribute this negative to both the two X and the 10. So we'll have 11 minus two X and minus 10 all over X plus five still greater than or equal to zero. Now, just combining like terms in the numerator, we have 11 minus 10 and 11 minus 10 is one. So we have a one minus two X over X plus five is greater than or equal to zero. Okay. The first step is done. Step two, for solving rational inequalities, we ask ourselves what values would cause the numerator and the denominator to equal zero. So we have got a variable in both the numerator and the denominator. So we can set both of those equal to zero and solve for X will set one minus two X equal to zero and we will set X plus five equal to zero. First solving for the one minus two X. I'm going to subtract one from both sides of the equation. And then we have negative two X equals negative one. Now I'm going to divide both sides by negative two. And on the left we have X and on the right negative one divided by negative two, that's a positive one half. And now for X plus five equals zero, I'm going to subtract five from both sides of the equation and we get X equals negative five. Well, those are two X values. And what are those telling us? Well, those are our boundaries. We draw a number line here from negative infinity to positive infinity. And our two boundaries are one half and negative five. I'm going to draw negative five in here first and then I'm going to draw one half in here. And what do these boundaries tell us? Well, the boundaries create intervals. We have three of them. One is from negative infinity to negative five. The next is from negative 5 to 1 half and the last one is from one half to infinity. And now for step three, we're going to choose test points from each of those three intervals from negative infinity to negative five. I'm going to choose negative six from negative 5 to 1 half. I'm going to choose zero and from one half to infinity. I'm going to choose one. Now we're going to fill each of those test points in four X in the original inequality. And we'll see where this is true. So we'll start off with negative six in our original inequality. We have 11 over. Not now, not X will put negative six in for X plus five is greater than or equal to two and simplifying our denominator on the left, negative six plus five equals negative one. So we have 11 divided by negative one is greater than or equal to two. Well, 11 divided by negative one that is negative 11. Is it true that negative 11 is greater than or equal to two? Nope, it's not. So that tells us that the interval from negative infinity to negative five is not true for this inequality. Now let's test out zero. So we still have 11 divided by and now not X but zero plus five is greater than or equal to two. Simplifying our denominator zero plus five is five. So we have 11, 11 5th. Excuse me, 11/5 is greater than or equal to two, 11 5th. That's two and 1/5. Is it true that two and 1/5 is greater than or equal to two? Yes, it is. That is true. So from negative 5 to 1 half, this inequality is true. Now let's test out one filling in one for X, we have 11 over not X but now one And then that's plus five is greater than or equal to two. And for our denominator one plus five is six. So we have 11/6 is greater than or equal to two. Well, 11/6, that's just one and 5/6 is one in greater than or equal to two. No, it's not. So that tells us this test point tells us that the interval from one half to positive infinity is not true for this inequality. So moving this X equals one half to the side for a second, we know that the interval is from negative 5 to 1 half. How do we express this an interval notation? Well, we can put negative five comma one half. But now are we talking about parentheses or brackets? Well, here is where it depends where we got the negative five and the one half from the um first thing we check is we see our original inequality was a greater than, or equal to. So the, or equal to part tells us that perhaps our boundaries can be included. But because these are rational inequalities, it depends whether the boundary came from the numerator or the denominator negative five that came from setting the denominator equal to zero. If we plug negative five and for X, we have negative five plus five and negative five plus five, that equals zero. And if our denominator equal zero, our fraction is undefined, we cannot have that. So even though this is a greater than, or equal to negative five gets a parentheses to say sorry. Negative five, you are not included. How about one half that came from setting our numerator equal to zero and setting the numerator equal to zero that can be included because we have that greater than or equal to zero. So one half gets brackets to say one half, you are included. You do get to come along for the ride. So where in our answer choices do we have parentheses? Negative five comma one half bracket. Well, that 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. 1/(x+2)≥3

Key Concepts

Rational Inequalities

Interval Notation

Critical Points

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