# Thread: [SOLVED] Rational inequality with an absolute value in the numerator

1. ## [SOLVED] Rational inequality with an absolute value in the numerator

Thanks!

2. Originally Posted by nihil.ad.rem
Hi. How would I go about solving this problem?

| x - 3 | / (x+4) is less than or equal to 5.

I'm not really sure how to isolate the absolute value expression.

Thanks!
hello,

$\frac{|x-3|}{x+4}\leq 5$ transform into:

$\frac{x-3}{x+4}\leq 5 ~ \wedge ~ x\geq 3~~\vee~~ \frac{-x+3}{x+4}\leq 5 ~ \wedge ~ x < 3$

Solve both inequalities separately. You should get as a final result:

$x < -4 ~~\vee~~x \geq -\frac{17}{6}$

3. Originally Posted by nihil.ad.rem
Hi. How would I go about solving this problem?

| x - 3 | / (x+4) is less than or equal to 5.

I'm not really sure how to isolate the absolute value expression.

Thanks!
I won't give the details, but another way would be to simply square the expression. $(|x|)^2 = x^2$. The downside is that you may have extra solutions, and instead of one inequality you have two:
$-25 \leq \frac{(x - 3)^2}{(x + 4)^2} \leq 25$.

However, despite the downsides I like the logic inherent in this method slightly better, even if it makes the problem more difficult to solve. And, of course, when you are done you get earboth's answer.

-Dan

### What if you have an absolute value in the numerator of a rational inequality? What will you do

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