Solve each rational inequality in Exercises 43–60 and graph th...

Question

Solve each rational inequality in Exercises 43–60 and graph the solution set on a real number line. Express each solution set in interval notation. (−x−3)/(x+2)≤0

Accepted Answer

Welcome back. I am so glad you're here. We are given the rational inequality negative x minus six, divided by x plus three is less than or equal to zero. We are asked to solve for this rational inequality and graph the solution on a number line. Well, the first step to solving irrational inequalities that we want to have one side equal to zero and yay it already is. So that steps done for us. The next step is we're going to find our boundaries. We do that by setting the numerator negative x minus six in the denominator, X plus three, both equal to zero. And then we'll solve for X. And that will give us our boundaries. So for the first one negative x minus six equals zero. Let's add six To both sides. Then we'll have negative x equals a positive six. Then we'll divide both sides by negative one and we have X equals negative 66 divided by negative one is negative six. There's one of our boundaries. Now let's find the other one will subtract three from both sides of X plus three equals zero. And then we'll have X equals negative three. So we've got our two boundaries. Now we can draw those on a number line. Our first boundary is going to be the negative six and the next one will be negative three. And now that we've got our two boundaries, we have three intervals. The first is from negative infinity to negative six. The next is from negative six to negative three. And the third intervals from negative three to positive infinity. And what we want to do is choose a test point in each of those three intervals and then we're going to plug the value of the test point in for X in our original inequality and see if this statement is true. If negative X minus six divided by X plus three is less than or equal to zero. So for the test point between negative infinity and negative six, I am going to choose negative seven. For the test point between negative six and negative three, I am going to choose a negative four. And for the test point between negative three and positive infinity I am going to choose zero. So now I'm going to plug each of those values in for X and see if the inequality is true. So we'll have a negative instead of X. It will be a negative seven minus six Over negative seven. Instead of X plus three is less than or equal to zero. Well, negative negative seven in the numerator is positive seven minus six over negative seven plus three. So seven minus six. In the numerator is one over negative seven plus three is negative four. So we have negative one. Fourth is less than or equal to zero. Is that true? Is negative 1/4 less than or equal to zero? Yes, it is. So for value from negative infinity to negative six our inequality is true. Now let's test negative four will have a negative negative four minus six over negative four plus three is less than or equal to zero. Negative negative four gives us a positive four. So four minus six over negative four plus three and taking four minus six. That's negative two over negative four plus three. That is negative one. So we have negative two over negative one which equals a positive two is positive two less than or equal to zero nope it is not therefore numbers between negative six and negative three are not true for inequality. Now let's pick numbers greater than negative three will choose zero. So we've got negative zero minus 6/ plus three is less than or equal to zero. While negative zero minus six that's zero minus six and that's just negative six. In the numerator and zero plus three is three. So we've got negative six thirds. That's negative two is less than or equal to zero. Is that true? Is negative two less than or equal to zero? Yes it is. So for values greater than negative three this inequality is true. What about negative six and negative three themselves? Well, for negative six we found that when we set the numerator equal to zero and let's take a look at our original inequality it says we're looking for when the left side is less than or equal to zero. Well when the numerator is equal to zero then it's zero. So six works. Excuse me. Negative six works. So it's included. How do we show that it's included? We draw a bracket on negative six and then we can draw a line going down to negative infinity to show that it does include all the numbers less than negative six as well. How about negative three? We found negative three. When we set the denominator equal to zero and when the denominator equals zero, well then the whole fraction is undefined and we can't have that so we can't have negative three. How do we show that we're not including negative three? We're going to put a parentheses at negative three and then draw our line going all the way up to toward positive infinity showing it does include the numbers greater than negative three. How do we indicate this? In interval notation? Will we put a parentheses negative infinity comma negative six. And then we're going to have a bracket to say including negative six and that's in union with. And now a parentheses to say not including negative three but all the numbers immediately after negative three comma all the way up to positive infinity. And now we look at our answer choices and this matches with answer choice. D Well done. We'll catch on the next one

Solve each rational inequality in Exercises 43–60 and graph the solution set on a real number line. Express each solution set in interval notation. (−x−3)/(x+2)≤0

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation and Graphing on the Number Line

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