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. 3/(x+5)>2

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 12 divided by the quantity of X plus seven is greater than six. Our answer choices are answer choice. A open parentheses, negative infinity to negative seven closed parentheses in union with open parentheses, five to infinity, closed parentheses. Answer choice B open parentheses, negative infinity to negative seven closed parentheses in union with open parentheses, negative five to infinity. Closed parentheses. Answer choice C open parentheses, negative seven to negative five closed parentheses and answer choice D open parentheses, negative seven to a positive five closed parentheses. All right. When we are trying to solve four inequalities, and we have a fraction on one side, we need to do a couple of things. We want to set one side equal to zero. And then we want to ensure that the other side has only one fraction. So we will begin by subtracting six from both sides so that the right hand side will be equal to zero on the left. We'll have 12 divided by X plus seven, then minus six. That's all. Now greater than six minus six is zero. We have zero on one side, but we have two things on the left, not just one single fraction. So we need to do the subtraction here when we're subtracting and we have a fraction, we need to have a common denominator. In order to subtract the common denominator will be X plus seven. So we multiply this six by X plus seven divided by X plus seven. So now we'll still have 12 divided by X plus seven. Now, that's minus six multiplied by the quantity of X plus seven in the numerator. And that's divided by X plus seven in the denominator still all greater than zero. Now, we have a common denominator and we can subtract our numerators. So we'll have 12 minus six multiplied by the quantity of X plus seven all divided by X plus seven and all greater than zero. Now, we can distribute this negative six to the X and the seven in parentheses. So we still have 12 and now negative six, multiplied by X is a negative six X and negative six multiplied by positive seven will be a negative 42. And that's all still divided by X plus seven and all still greater than zero. Now, we can combine the like terms in the numerator. We have 12 minus 42 12 minus 42 will be a negative 30. So we have negative 30 minus six X in the numerator all divided by X plus seven, all still greater than zero. And now we have finally met those conditions for which we can solve. For the following inequality, we have one fraction on the left and it's equal to zero on the right of the inequality. So now to solve for this inequality, we will set both the numerator and the denominator equal to zero. That will give us the te the end points of our intervals. So setting the numerator equal to zero, negative 30 minus six, X equals zero. I'll start by adding 30 to both sides. On the left. We have negative six X that equals on the right positive 30. Now we'll divide both sides by negative six. On the left. We have X and that is equal to on the right 30 divided by negative six is negative five. So one of our end points is negative five. Now we will set our denominator equal to zero X plus seven equals zero. Solving for X to get an end point, we subtract seven from both sides on the left. We have X equals on the right. Zero minus seven is negative seven. There's our other end point at negative seven. Now we will take these end points for our intervals and we will use them to test out our intervals and find out which ones are true. We'll draw a number line from negative infinity to positive infinity and writing down our end points, the first one going from negative infinity on the left that we encounter will be negative seven. And the next end point that we encounter will be negative five. And we have three intervals, one from negative infinity to negative seven. The next from negative seven to negative five and the third from negative five to positive infinity, we will choose a test point from each of those three intervals plug that test point in for X in our inequality and see if the inequality is true. If the inequality is true, that interval is included. If the inter, if the inequality is not true, that interval is not included. So for the test point from negative infinity to negative seven, I'm going to choose a negative eight. For the test point from negative seven to negative five, I'm going to choose negative six. And for the test point from negative five to positive infinity, I'm going to choose zero. So now plugging those test points in for X, we'll have for the test point negative eight, we'll have negative 30 minus six not multiplied by X but by negative eight all divided by not X but negative eight plus seven. And that's all supposed to be greater than zero. So simplifying in the numerator, negative six multiplied by negative eight will be a positive 48. So we'll have negative 30 plus 48 in the numerator and in the denominator negative eight plus seven will be a negative one. Simplifying further negative 30 plus 48 will be a positive 18. That's divided by negative one and 18, divided by negative one is negative 18 is negative 18, greater than zero. No, it is not. So this interval is not included. All right, let's test the next interval from negative seven to negative five. Using the test point negative six, we'll have a negative 30 minus six not multiplied by X but by negative six. And that's divided by not X but negative six plus seven. Supposedly all greater than zero. Simplifying in the numerator negative six multiplied by negative six is positive 36. So we have negative 30 plus all divided by negative six plus seven is going to be a positive one. Simplifying in the numerator, negative 30 plus 36 will be a positive six. Still divided by +16, divided by one is six is six greater than zero. Yes, it is. So this interval is true. Now we can test our final interval. So plugging zero in for X, we have negative 30 minus six, not multiplied by X but multiplied by zero and all divided by not X but zero plus seven, all supposedly greater than zero in the numerator simplifying negative six, multiplied by zero is zero. So that's just going to be a negative 30 in the numerator and in the denominator zero plus seven is positive seven. Well, we can keep that as the fraction negative 37th is that greater than zero? No, it's not, that's a negative number. So that one is not true. That interval does not work our interval for which this inequality is true is from negative seven to negative five. Now, how do we express that in interval notation? For interval notation, we use parentheses and brackets to indicate whether something is not included, which is with parentheses or included, which is with brackets. And we look at our original inequality and this one just says greater than it can't be equal to zero. So none of our test point or our end points, excuse me, none of our end points will be included because those made this equal to zero for the numerator and undefined for the denominator. So we cannot include those test points. So we're just going to use parentheses to say they're not included. So from, we start with the more negative one, the one closer to negative infinity, we say open parentheses from negative seven comma all the way up to negative five, but closing with parentheses because negative five is not included either. And that is our inequality expressed in interval notation. That's where this is true. We look at our answer choices and this matches with answer choice. See 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. 3/(x+5)>2

Key Concepts

Rational Inequalities

Domain Restrictions

Interval Testing and Sign Analysis

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