Solve each inequality. Give the solution set using interval no...

Question

Solve each inequality. Give the solution set using interval notation. 3/x-1 ≤ 5/x+3

Accepted Answer

Hello, everyone. In this video, we're going to be looking at this practice problem which states find the solution set of the following inequality and express in interval notation. And the inequality is four over X minus eight less than or equal to eight over X plus nine. So in this case, we're given four answer choices A B C and D, all of which are intervals with A and C having unions and B and D being standalone intervals. So in order to figure out which of these is correct, let's take a look at our inequality. So we have four over X -8 less than or equal to eight over X plus nine. And in order to solve this inequality, I need to isolate X and right now X is located on the left hand side and the right hand side of my inequality. So I'm going to move all my terms over to the left hand side by subtracting by eight over X plus nine on both sides. So my inequality becomes four over X minus eight minus eight over X plus nine less than or equal to zero. And now I have two fractions on my left hand side. And in order to combine them, I need them to have a common denominator, which means that I need to multiply four over X minus eight by the denominator X plus nine. So I need to multiply by X plus nine over X plus nine. And I need to multiply eight over X plus nine By X -8 over X -8. So when I do that, I have four multiplied by X plus nine over X plus nine, Multiplied by X -8 in my first fraction. And my second fraction becomes eight, multiplied by X -8 over X plus nine, Multiplied by X -8. Now, you can see that my fractions have the same denominator so I can combine them and this is still less than or equal to zero. So my fraction becomes four, go ahead and distribute this four. So four multiplied by X is for X four, multiplied by nine is 36. So four x plus 36, I can combine them now. So I have negative eight multiplied by X is negative eight, X, negative eight, multiplied by negative eight is positive 64. And that's all over the common denominator of X plus nine multiplied by X minus eight. And the fraction is still less than or equal to zero. I can combine like terms in my numerator leaving me with four, X minus eight, X is negative for X and 36 plus 64 is 100 over the denominator X plus nine, multiplied by X -8, still less than or equal to zero. And from here, you can see that we have all the excess on one side. And in order to find the solution set, I'm going to solve for X in the numerator and in the denominator separately and use those solutions to build intervals that I need to test. So if I sold for X in the numerator, I have negative four, X plus 100 is equal to zero to isolate X. I'm gonna subtract 100. So that leaves me with negative four, X is equal to negative 100 To completely isolate X, I'm going to divide by -4. So that leaves me with X is equal to negative 100 over negative four is positive 100 over four, which I can simplify to 25. So x equals 25 is my solution in the numerator. And if I look at the denominator, I have X plus nine is equal to zero, but I also have X minus eight is equal to zero. So if I solve for my first factor, I'm going to subtract nine, leaving me with X is equal to negative nine. And if I solve for my second factor I have X is equal to positive eight. So from here, you can see that I have three solutions for X which means that when I split my X axis Intervals, I'm going to have four intervals. So my first tick label -9 And then I have eight and finally 25, all the solutions for X. And my first interval is this region here from negative infinity to negative nine and then from negative nine to positive eight And then from positive 8 to positive 25 and finally from 25 to positive infinity. So at each of these intervals, I'm going to test values and see if my inequality is true. So if I start with my first interval, negative infinity to negative nine, I'm going to test X at negative 10. So I'm going to plug in negative 10 into my inequality. So I have negative four multiplied by negative 10 Plus over -10 - multiplied by negative 10 plus nine greater than Or less than or equal to zero. So negative four multiplied by negative 10 is positive 40 plus 100 over negative minus eight is negative multiplied by negative 10 plus nine, negative one less than or equal to zero. So 40 plus 100 is 1 Divided by -18, multiplied by -1 which is positive 18. So this is saying that a positive number is less than or equal to zero, which is false. So I can exclude this interval from my solution set. And if I look at my second interval, negative 9 to 8, I'm going to test X at zero. So I have negative four multiplied by zero plus over zero minus eight multiplied by zero plus nine, less than or equal to zero. So negative four multiplied by zero is zero. So my numerator is just 100 and that is over zero minus eight which is negative eight multiplied by zero plus nine, which is nine Less than or equal to zero. So I have 100 over negative eight Multiplied by nine, which is -72. So this 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 if I test my third interval eight to positive 25 I'm going to test X at 10. So I have negative four multiplied by 10 plus over 10 minus eight multiplied by plus nine less than or equal to zero. So negative four multiplied by 10 is negative 40 Plus 100 My Numerator. And that's over my denominator which is 10 -8, which is two And 10-plus 9, which is 19, less than or equal to zero. So negative 40 plus 100 is 60/2 multiplied by 19 Which is 57, Which is two times 19 is 38. And that is less than or equal to zero, which is not true. So I can exclude this interval from my solution set. And the final interval is 25 to positive infinity. And I will test x at 26. So I have negative four Multiplied by 26 plus Divided by 26 -8, Multiplied by 26 plus nine Less than or equal to zero. -4 times 26 is negative plus 100 in my numerator. And my denominator is 26 minus eight, which is 18 multiplied by 26 plus nine, just 35 less than or equal to zero. So my final fraction is negative 4/18 times 35 which is 630 less than or equal to zero. And in this case, this inequality saying that a negative number is less than zero, which is correct. So I can include This 4th interval in my solution set. So if we look at this number line that I drew, I don't want to include my first interval, but I can include my second interval and my fourth interval. But now I need to know if I should include my points. So I need to know if I should include negative nine eight and 25. So if I plug in X equals negative nine, I want to solve for those points to see if I should include them. I have negative four multiplied by negative nine plus 100 Over negative 9 -8 multiplied by - plus nine. And in this case, you can see that negative nine plus nine gives us zero. And if we have zero in the denominator, we get an undefined answer. So I know that I can't include -9 in my solution. The next point we want to test is X at eight. So if I plug in eight, I have negative four multiplied by eight plus over eight minus eight, multiplied by eight plus nine. And once again, you can see that b Denominator would become zero because I have 8 -8, which is zero. So I get an undefined answer. So I can't include eight in my solution set. And the final value we need to test is 25. So we have negative four Multiplied by plus 100 Over 25 - multiplied by 25 plus nine. And this is less than or equal to zero. So negative four multiplied by 25 is negative 100 plus Over 25 -8 which is 17 and plus nine. Just 34. In this case, I get negative 100 plus 100 is 0/17 times which is Is less than or equal to zero, which is true because we have zero equal to zero. So I can include x equals 25 in my solution set. So if we go back to this X axis, I would have open circles at negative nine and positive eight. And a close circle at 25. And if I wrote that in interval notation, I have parentheses negative nine to positive eight parentheses union with a bracket From to positive infinity. So this becomes my final answer for this practice problem. And if we look at the answer choices, you can see that answer choice C also has the same interval from a negative nine to positive eight union with 25 to positive infinity with - and eight having parentheses and 25 having a bracket. So hopefully this video helped and see you next time.

Solve each inequality. Give the solution set using interval notation. 3/x-1 ≤ 5/x+3

Key Concepts

Solving Rational Inequalities

Finding Critical Points and Domain Restrictions

Interval Notation for Solution Sets

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