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. (x - 8)/(x - 4) < 3

Accepted Answer

Hello, everyone. In this video, we're going to be going over this party's problem which states for the given inequality expression solve and write the solution set an interval notation. And the inequality expression is X minus 11 over X minus 13, less than two were given four answer choices for this problem A B C and D and all of the answer choices are some sort of interval. With the exception of D being no solution and A and B having unions between two intervals and C being a standalone interval from negative infinity to positive infinity, which is really all real numbers. So in order to figure out which of these answers, which of these answers is correct. Let's take a look at our inequality. So we have X -11 over X -13 less than two. So in order to solve this inequality, we want to solve for X. And in order to do that, I'm going to move on my terms to the left hand side, which means I need to move the two over to the left hand side. So I'll subtract both sides by two. So it leaves me with X -11. Over X -13 -2, less than zero. Now, I have two terms on the left hand side. But in order to combine them, I need them to have the same denominator, which means that I need my negative two, Took A, a denominator of X -13. So I'll multiply by X -13 over X -13. So when I do that, I can now combine my two terms. So I'll have X -11 minus two Multiplied by X -13 Over the common denominator, X -13 Still less than zero. And I can work on simplifying my numerator. So I have negative two multiplied by X is negative two, X, I still have the X minus 11 in front and negative two multiplied by negative 13 is Positive 26 and is still divided by X -13. And one last thing I could do to simplify my numerator is to combine like terms. So I have this X minus two X which is equal to negative X And then I have negative 11-plus 26 which is equal to positive 15. And that is still over X - And it's still less than zero. So from here, you can see that I've simplified my expression and all my terms are on the left hand side, but now X is still in the numerator and the denominator of my expression. So I'm gonna want to solve for X in the numerator and in the denominator. But let's look at the numerator first. So I have negative X plus 15 is equal to zero, I'll set it equal to zero. So to solve for X, I'll subtract 15 from both sides. So that leaves me with negative X is equal to negative And a divide by negative one to get rid of the negative in front of X. So that leaves me with X is equal to positive 15. That's one of my solutions for X. And if I look at the denominator, I have X -13 again, setting it equal to zero to isolate X, I'm going to add 13 to both sides. So it leaves me with X is equal to 13. So normally these would be my solutions. But because I have an inequality, these become the points of my intervals that I need to look at and test because I'm not trying to look for just one solution. I need to find the range of X values that fulfill this inequality. So when I look at a number line or the X-axis, I'm gonna draw a tick at 13 and another tick at 15. And from here, you can see that we have Three intervals that we're gonna want to test. There's x values from negative infinity to 13, X values from 13 to And from 15 to positive infinity. So I want to plug in values from those intervals and see if my inequality is still true. So starting with the first interval, negative infinity to 13, I'm going to plug in X at 10. So when I plug in 10, for X I get negative 10 plus 15 divided by negative 10 minus 13 is less than zero, negative 10 plus 15 is positive five divided by negative minus 13 which is negative 23 less than zero. This inequality saying a negative number is less than zero, which is true. So I can include this interval in my solution set. Then we can test the second interval which is 13-15. I'm going to plug in X equals 14. So when I plug in 14, I get negative 14 plus 15/ minus 13 is less than zero. So negative 14 plus 15 is 1/14 minus 13 which is also one that's less than 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. And the final interval we need to test is 15 to positive infinity. So I'm going to plug in x equals 16. When I do that, I have negative 16 plus 15/16 -13 is less than zero. Negative 16 plus 15 is negative one Over 16 -13 which is three, less than zero. So this is saying a negative number is less than zero, which is true. So I can include this interval in my solution set. So if we look at this number line, I can include 13 To negative infinity or negative infinity to 13 and 15 to positive infinity. And notice how I did an open circle because in my inequality, I have a less than and not equal to. So if I were to write this interval in interval notation, I have parentheses, negative infinity to 13 union with Parentheses again, 15 to positive infinity would become my final answer for this practice problem. And if we look at the answer choices, you can see that answer choice B Also has the same interval from negative infinity to positive 13 Union with 15 to positive infinity. 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. (x - 8)/(x - 4) < 3

Key Concepts

Rational Inequalities

Interval Notation

Critical Points

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