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. (x-3)/(x+5)≤0

Accepted Answer

Hello everyone. In this video, we're going to be looking at this practice problem which states solve the following inequality and express the answer in interval notation. And the inequality is X minus 14 divided by X plus 22 less than or equal to zero. And we're given four answer choices A B C and D and you can see that all of the answer choices are intervals with varying interval notation, parentheses or brackets and A and D have unions. Well, B and C are standalone intervals. So let's go ahead and look at our inequality to see which of these are correct. So our inequality is X -14 divided by X-plus 22. And in this case, you can see that to solve this inequality and isolate X, we have X in both the numerator and denominator, which means we need to consider X in both of these situations. So if we solve for X in the numerator, I have X -14 is equal to zero. If I add a fortune to both sides, I get X is equal to positive 14. If I solve X in the denominator, I have X plus 22 is equal to zero in order to solve for X also tracked 22. So that leaves me with X is equal to negative 22. And you might be thinking these are the final solutions but these are just the indicators of what intervals we should be looking at for the solution set. So if I draw a number line or the X axis and I put, we'll call this negative 22 this tick positive 14, I need to be looking at the intervals that these brakes cause. So you can see that I have this interval here from negative infinity to negative 22 then a second interval from negative 22 to positive 14 And then my third interval from positive 14 to positive infinity. So now that I have these intervals, I need to test values in each of those intervals and see If my inequality is true in those intervals. So if I look at negative infinity to negative 22, I can try the value U X equals negative 23 since negative 23 is in that interval. So I'll plug in negative 23 to my inequality. So I have negative 23 minus 14 divided by negative 23 plus 22 less than or equal to zero. So this becomes negative 23 minus 14 is negative divided by negative 23 plus 22 which is negative one. So two negatives make a positive so I'm left with positive divided by one is less than or equal to zero. And in this case, that is not true, I have a positive number On the left hand side. So it's not less than zero. So I know that this interval is not included in my solution set. So I can look at the second interval which is negative two positive 14. And I'm going to go ahead and test X equals zero. So if I plug in zero, I have zero minus 14 divided by zero plus less than or equal to zero. So zero minus 14 is negative 14 and zero plus 22 is positive 22 less than or equal to zero. And in this case, I have a negative number being less than or equal to zero, which is true. So I can include the interval negative 22 to positive in my solution set. And the final interval I need to test is from to positive infinity. I'm gonna go ahead and try X equals 15. And when I try 15, I have 15 minus 14 divided by 15 plus 22 is less than or equal to 0. 15 minus 14 is one divided by 15 plus 22 which is 37 Less than or equal to zero. And in this case, I have a positive number being less than zero, which is not true. So I don't include this interval, Which means that the only interval that I'm including in my solution set is from negative 22 two positive 14. So a final check is to see if we include the endpoints of the interval -22 or 14. So if I plug in negative 20 to 4 X I have negative 22 minus 14 divided by negative plus 22 less than or equal to zero. Well, in this case, you can see that negative 22 plus 22 is zero. So that gives me an undefined answer. So I don't want to include negative 22 in my interval. Which means if I draw this on a number line, it would be an open circle. Now we can test the other endpoint which is X at 14. And if I plug in 14, I have 14 minus 14 divided by 14 plus 22 is less than or equal to zero. And this becomes zero divided by 14 plus 22 which is 36 is less than or equal to zero and zero divided by 36 is in fact zero. So this is a true inequality Which means that I need to include X equals 14 in my solution set. So if I draw on a number line, it would be a closed circle. And if I wrote it as an interval or an interval notation, I need to exclude -22. So parentheses negative 22 up to positive 14 and I need to include 14. So I do a bracket. So this becomes my solution set for this problem. And if we look at the answer choices, you can see that answer choice C also has negative to positive 14 with the bracket On the 14. So hopefully this video helped and see you next time.

Solve each rational inequality. Give the solution set in interval notation. See Examples 8 and 9. (x-3)/(x+5)≤0

Key Concepts

Rational Inequalities

Interval Notation

Critical Points

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