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−4)/(x+3) > 0

Accepted Answer

Welcome back. I am so glad you're here. We are given the rational inequality, X minus two, divided by X plus six is all greater than zero. We are to solve this rational inequality and graph it on a number line. Well when we are solving rational inequalities, that first step is to get one side equal to zero and ye the right hand side is already equal to zero. That steps done next. We are going to find our boundaries. We do that by setting our numerator X minus two and our denominator, X plus six both equal to zero. Then we're going to solve for X. And those will be our boundaries. So for the first one x minus two equals zero will add two to both sides. And that will give us one boundary at X equals to four. X plus six equals zero will subtract six from both sides and then we'll have another boundary at X equals negative six. Now we can plot those on a number line. The first one will be negative six. And the next one will be to. Now the next step is that we are going to identify our intervals. Our first interval is from negative infinity to negative six. The second interval is from negative 6 to 2 and the third interval is from two to positive infinity. Now that we've identified those intervals, we're going to choose test points for each of those. So from negative infinity to negative six. I'm going to choose negative seven. From negative 6 to 2. I'm going to choose zero and from two to positive infinity I'm going to choose three. And now what we'll do with each of those test points is we will plug those in for X. In our original inequality. And we will see if this inequality remains true that the left hand side will be greater than zero. If it's true then that interval is included for our inequality. So let's start out by testing negative seven will have instead of x minus two is negative seven minus two over instead of X plus six. It's negative seven plus six is all greater than zero, negative seven minus two is negative nine and negative seven plus six is going to be negative one. So negative nine divided by negative one. That's positive, nine is nine greater than zero. Yes it is. So for values from negative infinity to negative in six, negative six this inequality is true. Now let's test zero will test the next interval we have zero minus two. In the numerator over zero plus six is greater than minus two is negative two and zero plus six is a positive six. So we've got negative 2/6 or negative one third is negative one third greater than zero. No it is not. So for values from negative 6 to 2, this inequality is not true. Now let's test three we have three minus two. In the numerator over three plus six in the denominator is greater than 03 minus two is 1/3 plus six is nine. So we have 1/9 is greater than zero. Is that true? Yes. 1/9 is a positive number. It is greater than zero. So this interval or this inequality is true for the interval two to positive infinity. What about -6? And to themselves? Well we found negative six. When we set the denominator equal to zero and when the denominator is equal to zero, the whole fraction is undefined and we can't have that. So negative six does not work. We denote that negative six cannot be included by putting a parentheses and then we can draw a line from negative six down toward negative infinity to show that the numbers from negative infinity to negative six are included. How about two? Well we found two when we set the numerator equal to zero which would make the entire fraction equal to zero. But this inequality is just setting it greater than zero. So two cannot be included either. We again to is not included by putting a parentheses there and then we do draw a line going toward positive infinity showing that the numbers greater than two are included for our inequality. How do we denote this in interval notation? Well, we'll start by putting parentheses negative infinity comma. All the way up to negative six parentheses showing negative six is not included in union with parentheses starting at positive two comma. All the way up to positive infinity parentheses and that is it. In interval notation. We look at our answer choices and this matches with answer choice C. Well done, we'll catch you 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−4)/(x+3) > 0

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation and Graphing on a Number Line

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