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. (2x + 3)/(x - 5) ≤ 0

Accepted Answer

Hello everyone. In this video, we're going to be looking at this practice problem which states for the given inequality expression solve and write the solution set in interval notation. And the inequality expression is four X plus seven over X minus 13 less than or equal to zero. And were given four answer choices for this problem. All of which are intervals except for the last answer, choice answer choice D which is no solution. So to figure out which of these answer choices is correct, let's look at our inequality. So we have four X plus seven Over X -13 Less than or equal to zero. In this case, we already have all of our terms on the left hand side. But we have X in both the numerator and denominator. So because to solve this inequality, we need to isolate X, we're going to need to solve for X in the numerator and the denominator. So if we solve for X in the numerator, we have four X plus seven Is equal to the right hand side of the inequality which in this case is zero. So if I solve for X, I'm gonna subtract by seven. So it leaves me with four, X is equal to negative seven And to isolate X. I'm going to divide by four. So that leaves me with X is equal to negative 7/4. So that's X in my numerator. If I look at the denominator, I have X -13 is equal to zero. Again, I'm going to add 13 to both sides. So that leaves me with X is equal to 13. And normally this would be my solutions for X. But in this case, we have an inequality and not necessarily an equal, equal to equation. So because we have inequality, these points become the indicators of what intervals I need to test to see what range of X values satisfy this inequality. So I'm going to draw a number line or the X axis and I'm going to put ticks at my solutions for X. So my first tick will be at negative 7/4 and my second tick will be at 13. And These are the solutions for X. But these are also where I need to look at for intervals. So if we look at this number line, you can see that I have an interval from negative infinity to negative 7/4 And then have, and then I have another interval from -7/4 to positive 13. And my final interval is from 13, 2 infinity. So in order to find our solution set or the range of X values that satisfy this inequality, we need to test values in each of these intervals. So starting with the first interval, negative infinity to negative 7/4 I'm going to try X negative two. So if I plug in negative two into my inequality, I have four multiplied by negative two plus seven over X Or -2 - Less than or equal to zero. So if I start to simplify my numerator four multiplied by negative two is negative eight plus seven over negative two minus 13, which is negative 15, less than or equal to zero, negative eight plus seven is negative one over negative 15. Two negatives make a positive. So I have 1/15 is less than or equal to zero. So this is saying that a positive number is less than zero, which is not true. So I can exclude this interval from my solution set. Looking at my next interval, negative 7/4 to 13, I'm going to test X at X equals two. So I have four multiplied by two plus 7/2 minus 13. Less than or equal to 04, multiplied by two is eight plus 7/2 minus 13, which is negative 11 less than or equal to 08 plus seven is 15 over negative 11 less than or equal to zero. So this inequality is saying that a negative number is less than or equal to zero, which is true. So that means that I can include this interval in my solution set. And the final interval I need to test is from to infinity. So I'm gonna try x 14. So I have four multiplied by 14 plus seven Over 14 -13 Less than or equal to zero. Four times 14 is equal to 56 plus 7/14 minus 13, which is one less than or equal to zero. So 56 plus seven is equal to 63 less than or equal to zero. And this is saying that a positive number is less than or equal to zero, which is not true. So I can exclude this interval from my solution set. So if I go back to this number line that I drew, the only interval Where the inequality was true was between -7/4 and 13. So if I were to visualize this solution set on a number line, I would have a line from negative 7/4 to positive 13. But now I need to know if I should include the endpoints negative 7/4 and 13. Well, in this case, because I have the less than or equal to, I can include the endpoints as long as they don't make my expression undefined. So because I'm working with a fraction, the only time where this would become undefined is if my denominator X - is equal to zero. So if I solve for X I get that, when X equals 13, my solution is undefined, which means that X cannot equal 13, But it can equal -7/4. So I can do a close circle at -7/4. But at 13, it needs to be an open circle. And if I write this interval in interval notation, I would have a bracket at negative 7/4 up to 13 with a parentheses. And this becomes my final answer for this problem. If we look at the answer choices, you can see the answer choice C has the same interval from negative 7/4 to positive 13 with a bracket on negative 7/4 and a parentheses at 13. So hopefully this video helped and see you next time.

Solve each rational inequality. Give the solution set in interval notation. See Examples 4 and 5. (2x + 3)/(x - 5) ≤ 0

Key Concepts

Rational Inequalities

Interval Notation

Sign Analysis

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