Solve each rational inequality. Give the solution set in interval...

Question

Solve each rational inequality. Give the solution set in interval notation. See Examples 4 and 5. 1 /{x - 1} < 1 /{ x + 1}

Accepted Answer

Welcome back. I am so glad you're here. We are asked to provide the solution set for the given rational inequality in interval notation. Our given rational inequality is on the left, we have the fraction one divided by X minus 36. And that's less than on the right. We have the fraction one divided by X plus 36. Our answer choices are answer choice. A, the solution set open bracket negative 36 to 36 close parentheses. Answer choice B the solution set open parentheses negative 36 to 36 close bracket. Answer choice C the solution set open bracket negative 36 to 36. Close bracket hand answer choice D the solution set open parentheses negative 36 to 36 closed parentheses. All right. When we are solving for a rational inequality, we need to have two conditions met first, we need to have a zero on one side and then one single fraction on the other. So we can subtract one divided by X plus 36 from both sides of the inequality. So that on the right, it will be zero. Yay. But on the left, we'll have one divided by X minus minus one, divided by X plus 36 still less than zero. So we don't have the one single fraction on the left anymore. We do need to do the subtraction. And in order to subtract when we have fractions, we need to have a common denominator. The common denominator here is just going to be these two factors, the quantity of X minus 36 multiplied by the quantity of X plus 36. So how do we get there? Well, in order to get our first fraction to have that denominator, we need to multiply it by the denominator of the second fraction. So we're going to multiply that by the quantity of X plus 36 divided by X plus 36. And then to get that common denominator for the second fraction, we need to multiply that by the denominator of the first fraction. So multiplying it by the quantity of X minus 36 divided by X minus 36. And so then multiplying the numerators from that first fraction, one multiplied by X plus 36 is X plus 36. Then we need to bring over the subtraction sign and one multiplied by the quantity of X minus 36 in the numerators is just going to be the quantity of X minus 36. Noting this is all still less than zero. And now we can distribute this negative the subtraction to the numerator from the second fraction the X minus 36. So we can bring over X plus 36 then negative multiplied by X is a negative X negative multiplied by negative 36 is a positive 36 all divided by those two factors. X minus 36 multiplied by X plus 36. All still less than zero. And now we can combine like terms in the numerator X minus X. Well, those are canceled out. Great. A positive 36 plus 36 is a positive 72. So the numerator is 72. The denominators we have or the denominator is the quantities of X minus 36 multiplied by the quantity of X plus 36 all less than zero. And now we've met those two conditions where we have a single fraction on the left and zero on the other side. So now we can start, what do we do here? Well, what we need to do is anywhere there's an X, we're going to be setting our numerator or denominator equal to zero if there's an X and we're finding our end points. Well, there are no zeros in the numerator. So setting our denominators equal, uh our denominators factors equal to zero. We get X minus 36 equals zero and X plus 36 equals zero. Solving for X for the first one, we add 36 to both sides and we get X equals positive 36. For the second one, we subtract 36 from both sides. And we get X equals a negative 36. And so we have two end points of 36 negative 36. What do we do with those end end points? We can set up a number line from negative infinity to positive infinity. And we can note our end points. The first one from negative infinity is negative 36. And then the next one is positive 36. And we now have three intervals. The first interval is from negative infinity to negative 36. The second interval is from negative 36 to positive 36. The third interval is from positive 36 to infinity. What we're going to do is select a test point from each of those three intervals. We're going to plug the text test point in for X in our inequality. And we're going to see if it's true. If the inequality is true for the test point, then that interval is true for the inequality. So from negative infinity to negative 36 I'm going to choose the test point negative 40. You don't have to from negative 36 to positive 36. I'm going to choose the test 360.0 and from positive 36 to infinity, I'm going to choose the test 360.40. So now plugging those in for X and the inequalities. So we'll have 72 divided by not X but negative minus 36 multiplied by the quantity of not X but negative 40 plus 36 all supposedly less than zero. Simplifying in the denominator would keep that 72 in the numerator negative 40 minus 36 is a negative negative 40 plus 36 is a negative four. And doing the multiplication in the denominator negative 76 multiplied by negative four is a positive 304. So we have 72 divided by 304. Is that less than zero? No, it is not. So that inter that interval is not true for our inequality. Now, let's test for zero, we still have 72 in the numerator. And then in the denominator instead of X, we have zero minus 36 multiplied by not X but zero plus 36 supposedly less than zero doing the addition and subtraction in the numerator. Keeping 72 in the num the addition and subtraction in the denominator keeping 72 in the numerator zero minus 36 is negative 36 0 plus 36 is positive multiplying negative 36 multiplied by is a negative 1296. So we have 72 divided by negative 1296. Without simplifying that we can see that that's a negative number. So is that less than zero? Yes, it is. So that interval is true for our inequality. Now we can do the final interval with the test 0.40. So we're taking 72 divided by the quantity of 40 minus multiplied by the quantity of plus 36. All supposedly less than zero. Keeping 72 in the numerator, 40 minus 36 is a positive four and plus 36 is a positive 76 4 multiplied by 76 is 304. So we have 72 divided by 304 which is not less than zero. So the only interval for which this inequality is true is from negative 36 to 36. But does this include negative 36 36? No, it doesn't because this inequality is for less than and solving for negative 36 positive 36 would put our denominators equal to zero. And if our denominators equal zero, that's undefined. So no, the negative 36 36 are not included. How do we say they're not included? We put an open parentheses to say it's not included. The negative 36 comma 36 closed parentheses to say it's not included. And that's how we express this inequality in interval notation. We look at our answer choices and this 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 4 and 5. 1 /{x - 1} < 1 /{ x + 1}

Key Concepts

Rational Inequalities

Interval Notation

Critical Points

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