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. (3x + 7)/(x - 3) ≤ 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 five X plus six over X minus 17, less than or equal to zero. And were given four answer choices for this practice problem A B C and D, all of which are intervals with C having a union and D is actually not really an interval, it's no solution. So to figure out which of these answer choices is correct, let's take a look at our inequality. So for inequality, we have five X plus six over X minus 17 less than or equal to zero. So in this inequality, all of our terms are already on the left hand side, but we have a fraction and X is present in both the numerator and the denominator. So in order to find the solutions that we need to solve for X in the numerator and in the denominator. So looking at the numerator first, I have five X plus six and I'm going to set it equal to zero because that is what's on the right hand side of my inequality. So to solve for X, I'm gonna subtract six from both sides, leaving me with five, X is equal to negative six. And to completely isolate X, I need to divide by five on both sides. So I have X is equal to negative 6/5. And that's my solution for the numerator. And now I can look at the denominator Where I have X -17. And again, I'm going to set it equal to zero. So to solve for X, I'm going to add 17 to both sides. So that leaves me with X is equal to 17. So this is two solutions for X and normally these would be my answers. But because I have an inequality, these become the points where I need to look at intervals for solutions of X and see if my inequality is true at those intervals because of the inequality, I need to find the range of X values that satisfy B inequality. So if I draw a number line, I'm gonna put ticks for the solutions of X. So I have a tick at negative 6/5 and another tick at 17. And from here, you can see that I need to look at three intervals, the first one negative infinity to negative 6/5 the second one negative 6/5 to positive 17 And the last one from 17 to positive infinity. So I need to test X values in these intervals. To see if my inequality is true. So looking at the first interval, negative infinity to negative 6/5, I'm going to plug in X equals -2. So if I plug in X equals negative two, I have five multiplied by negative two plus six over negative two minus 17, Less than or equal to zero five, multiplied by -2 is negative 10 plus six over negative two minus 17 is negative 19 less than or equal to zero, negative 10 plus six is negative four over negative 19. Still two negatives make a positive. So I have 4/19 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. So it's not a solution. And my next interval is negative 6/5 17. And I'm going to test X at zero. So I have five multiplied by zero plus 6/0 minus 17. Less than or equal to 05, multiplied by zero is zero. So my numerator is just six and my denominator zero minus 17 is negative 17. So the less than or equal to zero and this inequality is saying that a negative number is less than or equal to zero, which is true. So I can include this interval in my solution set. And the last interval I need to test is 17 to positive infinity. So I'm going to test x at 20. So I have five multiplied by 20 plus 6/20 minus 17, less than or equal to zero. So five multiplied by 20 is 100 plus 6/20 minus 17, which is three less than or equal to 0. 100 plus six is 106 still divided by three 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 also exclude this interval from my solution set, which means that the only interval that caused my inequality to be true was from negative 6 5th to 17. So on a number line, It would be a line from -6/5 to 17. And the next question is if I should include the endpoints, negative 6/5 and 17. Well, because I have a less than or equal to, I can include the endpoints as long as they cause my expression to give me a valid answer. So if I plug in negative 6/5 into my expression, I get a valid answer. And the only case where I wouldn't get a valid answer is if my denominator is equal to zero and that happens when I plug in X equals 17. So when x equals 17, my denominator is equal to zero, which is no good Because that is, that gives me an undefined answer. So I need to make sure to not include 17 in my solution set. So if I were to visualize the solution on a number line, it would be a close circle from negative 6/5 up to 17 with an open circle. And if I write that in interval notation, I would have a bracket at negative 6/5. And That would go up to 17 with a parentheses because I need to exclude 17. So this becomes my final answer for this practice problem. And you can see that answer choice. A also has the same Interval with the bracket on the negative 6/5 and the parentheses on the positive 17. 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. (3x + 7)/(x - 3) ≤ 0

Key Concepts

Rational Inequalities

Interval Notation

Sign Analysis

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