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.
7/(x+2)≥1/(x+2)

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 six divided by X plus 13 which is greater than or equal to four divided by X plus 13. And our answer choices are a from negative infinity to negative 13 in union with 13 to infinity be from negative 13 to 13. C from negative 13 to infinity with brackets closing both of those and D from negative 13 to infinity with parentheses closing both of those. Alright. Well, we recall from previous lessons that the first step in solving a rational inequality is to set one side equal to zero. So to do that, I'm going to subtract four over X plus from both sides of the inequality. So on the left, we will have six over X plus minus four over X plus and that's still greater than or equal to. But on the right hand side, four over X plus 13 minus four over X plus 13, that's zero. Alright. And still part of this first step, we want to ensure that on the side that has the fractions that, that is combined to only a single fraction. So in order to subtract two fractions, we need a common denominator and we have one yay, our eight X plus 13 is our common denominator. So that means we can go ahead and subtract our numerator six minus four equals two. So we have two over X plus is greater than or equal to zero. Step two for solving rational inequalities is to ask ourselves what values would cause the numerator and or the denominator to equal zero. And in this case, only the denominator has a variable. So we set the denominator X plus 13 equal to zero. We're going to solve for X and see what values would make our denominator equal zero. So for X plus 13 equals zero, I'm going to subtract 13 from both sides of the equation and we get X equals negative 13. So what does this mean? Well, X equals negative 13 gives us our boundary. So we draw a number line going from negative infinity to positive infinity and we have a boundary at negative 13. And now what we're going to do, we've created two intervals, one from negative infinity to negative 13 and the other from negative 13 to positive infinity. And we'll choose a test point from each of those two intervals plug that test point in for X in the original inequality and see for which interval it is true. So from the interval from negative infinity to negative 13. I'm going to choose the test point of negative 14 for the interval from negative 13 to infinity. I'm going to choose the test point of zero. And now we plug those into the original inequality for X plugging in negative 14, we have six over not X but now negative 14 still plus 13 is greater than or equal to now four over not X but negative 14 still plus 13. Now simplifying those two denominators negative 14 plus 13 is negative one. So we have six over negative one is greater than or equal to four over negative one. Now simplifying six divided by negative one is negative six and four divided by negative one is negative four. So is it true that negative six is greater than or equal to negative four? No, it's not, not true. So that tells us our test point tells us that for the values from negative infinity to negative 13, that interval is not true for our inequality. Now we can test zero will have six over not X but zero plus 13 is greater than or equal to four over not X but zero plus 13. Simplifying our denominators, zero plus 13 is 13. So we have 6/13 C is greater than 4/13. So is it true that 6/13 is greater than or equal to 4/13? Yes, it is. And so that tells us that this inequality is true for the interval from negative 13 to positive infinity. Well, how do we express that in interval notation? Well, we can have negative 13 comma infinity. And now are we going to close these with parentheses or brackets? Well, parentheses means not including, and we're always going to close infinity with the parentheses to say we'll never quite obtain infinity. But what about negative 13 is that included? Well, there's two things we have to check here. First, we check our original inequality and we see that it's greater than, or equal to. So the or equal to usually tells us that a value is included. However, we're dealing with rational inequalities and where did that? Negative 13 come from? It came from checking our denominator and setting that equal to zero. If we plug negative 13 in for X, we have negative 13 plus 13, which is zero and that would give us an undefined fraction and we cannot have that. So sorry, negative 13, you're not included. You would make this undefined. So you get a parentheses to say not included. Alright, looking at our answer choices for from negative 13 to positive infinity. But using parentheses 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. 7/(x+2)≥1/(x+2)

Key Concepts

Rational Inequalities

Interval Notation

Critical Points

Watch next