Problem

Solve each inequality, and graph the solution set…

05:53

Problem 41 Hard Difficulty

Solve each inequality, and graph the solution set. See Examples 5 and $6 .$
$$
\frac{x-8}{x-4}<3
$$

Answer

$$
(-\infty, 2) \cup(4, \infty)
$$

Related Courses

Precalculus

Algebra

Beginning and Intermediate Algebra

Chapter 11

Quadratic Equations, Inequalities, and Functions

Section 8

Polynomial and Rational Inequalities

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Video Transcript

All right, So we have X minus eight over X minus four is less than three. So we want to find the solution set that makes this true. The first thing we have to have is we have to have on inequality that's being compared to zero. So the first thing I'm gonna do here, this subtract three from both sides, so we get X ministate over. X minus four minus three is less than zero. So now we have the zero on the right side, everything else on the left. The next thing that has to happen then is we have to have a common denominator to be able to subtract these two expressions. That common denominator is gonna have to be X minus four. So I'm gonna have to take the three and multiply it by x mass for over X minus four. And when we do that, we will get X Men ISI over X minus four minus three X minus 12 over X minus four. And that's all less than zero. So now I can start subtracting X minus three x I mean negative two x negative eight minus negative 12 is the same as saying negative eight plus 12. So you get plus four over X minus four and that is less than zero. Now, at this point in the problem, we want to take both the numerator and the denominator and set him equal to zero. So we will get negative. Two X plus four equals zero. An X minus four equals zero. We want to solve each of these for acts. Subtract four from each side. Negative two X equals negative. Four. When I divide by negative too, I get X equals two. And here we at affordable size. We get X equals for so those are my two routes. Now the thing about these routes is they represent different things. When I plug four in for acts, it would give me a zero Omar denominators. This four represents where the function would be undefined, so it can't be part of our solution set. And in this case, when X is equal to two, that indicates where the fraction will be equal to zero. But I don't want to be equal to zero. I only want to be less than zero. So too will also not be part of our solutions said in this particular case. So where is our solution Set? Wow, if we think about our number line, I know the two in the four cannot be part of our solution set, but those marked the end points of the intervals that are part of our solution set. So our interval, our solutions, that will either be the interval to the left of two. Interval between two. And four or the interval to the right of four. We won't know until we start testing some values. So I'm gonna pick something to the left of to, like, zero. A nice imager between two and four that's gonna be convenient would be three and something to the rate of four. We'll pick five. So those are the three values. I'm gonna use the test to see what interval make up our solution set. So if X is equal to zero, then we're gonna have zero minus eight were zero minus four. And if it checks, it will be less than three. Well, that's gonna be negative. Eight over negative four, which is two and two is less than three. So that indicates that those values to the left of two must be part of our solution Set. When we test X equals three, we get three minus eight over three, minus four, and I want to see if that's less than three. Well, that will be negative. Five over negative one, which is 55 is definitely not smaller than three. So that interval is not part of our solution set. So next, then we're gonna check X equals five. This will give us five minus eight over five, minus four, and that's going to give us negative 3/1, which is actually less than three. So that tells me that that interval checks So what is it that we know? We know our solution set is going to be the interval from negative infinity to To or the interval from Four to Infinity. That right there is our solutions that

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