# Thread: Rational Inequality With Absolute Value Confusion

1. ## Rational Inequality With Absolute Value Confusion

Here's the problem I'm confused with: |(3x-9) / (x-7)| < 1

From what I understand this means |(3x-9) / (x-7)| is greater than -1 and less than 1. Therefore,

-1 < (3x-9) / (x-7) < 1

And there are two cases. One where x-7 is positive and one where x-7 is negative. Therefore,

x-7 > 0 is x > 7 and x-7 < 0 is x < 7

In the first case I multiply the inequality by x-7 which gives me -x+7 < 3x-9 < x-7

Then I divide this up into -x+7 < 3x-9 AND 3x-9 < x-7

After solving the first part, I get 4 < x. After solving the second, I get x < 1.

This is where the confusion starts, unless I already messed something up at this point. X can't be greater than 4 AND less than 1. Therefore, this is not a solution, I believe.

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In the second case x-7 is negative, so when I multiply the inequality by it I reverse the signs. And I'm assuming I'm going to get the same answer, except with the inequalities reversed, so: 4 > x AND x > 1. In my head, this makes sense because X can be less than 4 while being greater than 1, right? So this seems good at first. Now this is where the real confusion sets in. Am I supposed to then compare 1 < X < 4 with X < 7? If so, that seems to check out as well. I can think of a number that is greater than 1, less than 4, and less than 7. But wouldn't that number make the denominator less than zero, which is unacceptable?

The bottom line is I'm not clear as to what I'm supposed to do. How do I compare X > 7 with 4 < X AND X < 1, then compare X < 7 with 4 > X AND X > 1... What I'm trying to say is I don't know how to use the conclusions to find the solution...

Apparently, the solution is supposed to be (7, infinite)... But I'm not sure how one gets that.

Any help would be much appreciated. Thank you.

2. ## Re: Rational Inequality With Absolute Value Confusion

you're on the right track. I think an algebra error threw you off.

$-1 < \dfrac{3x-9}{x-7} < 1$

case 1: $x > 7$

$7-x < 3x-9 < x-7$

$16 < 4x\text{ and } 2x < 2$

$4<x \text{ and }x < 1$

well these are clearly incompatible results so we dismiss them

case 2: $x < 7$

$-1<\dfrac{3x-9}{7-x}<1$

$x-7 < 3x - 9 < 7-x$

$2 < 2x \text{ and } 4x < 16$

$1 < x < 4$

This is both consistent on both sides of the inequality and consistent with the assumption of $x < 7$

I just prefer to do the algebra with the denominator positive. It helps keep things clearer.

3. ## Re: Rational Inequality With Absolute Value Confusion

\displaystyle \begin{align*} \left| \frac{3\,x - 9}{x - 7} \right| &< 1 \\ \left| \frac{3\,x - 21 + 12}{x - 7} \right| &< 1 \\ \left| \frac{3\left( x - 7 \right) + 12}{x - 7} \right| &< 1 \\ \left| \frac{3 \left( x - 7 \right) }{x - 7} + \frac{12}{x - 7} \right| &< 1 \\ \left| 3 + \frac{12}{x - 7} \right| &< 1 \\ -1 < 3 + \frac{12}{x - 7} &< 1 \\ -4 < \frac{12}{x - 7} &< -2 \\ -\frac{1}{2} < \frac{x - 7}{12} &< -\frac{1}{4} \\ -6 < x - 7 &< -3 \\ 1 < x &< 4 \end{align*}