Inequalities with absolute values and fractions

ozfingwe · #1 $Old$ November 2nd, 2009, 10:51 AM

hi,

I am facing a couple of issues with two inequalities, would appreciate help here.

1) |2x + 3| ≤ 3x − 5

the first case would be 2x + 3 ≤ 3x - 5, solves easily, but no approach to the second case makes sense to me:
-2x - 3 ≤ 3x - 5 ??
2x + 3 >= -3x + 5 ??
=> x >= 2/5 but that doesn't even come close to fitting the original inequality?

If I square them, I get
4x^2 + 12x + 9 <= 9x^2 - 30x + 25, solving that with the usual formula I end up with x1 = 2 and x2 = 8/5, which is also somewhat strange.

2) 3 / | x - 9 | > 2 / (x + 2)
The essential questions here are how many cases I have to distinguish between and if anyone could be so kind to formulate a general rule when the inequality sign changes direction and the signs on one (or both?) sides of the inequality change from + > - and vice versa.

Thank you.
Oz.

earboth · #2 $Old$ November 2nd, 2009, 02:31 PM

Quote:

Originally Posted by ozfingwe $View Post$

hi,

I am facing a couple of issues with two inequalities, would appreciate help here.

1) |2x + 3| ≤ 3x − 5

...

Use the definition of the absolute value:

$|2x+3|=\left\{\begin{array}{l}2x+3\ if\ 2x+3 \geq 0~\implies x\geq -\dfrac32 \\ -(2x+3)\ if\ 2x+3 < 0~\implies x < -\dfrac32 \end{array}\right.$

So you have to solve 2 different equations:

$2x+3 \leq 3x-5~\wedge~x\geq -\dfrac32$

$\boxed{8 \leq x}~\wedge~x\geq -\dfrac32$
-------------------------------------------------------------------------
$-(2x+3) \leq 3x-5~\wedge~x < -\dfrac32$

$2 \leq 5x~\wedge~x < -\dfrac32$

$\dfrac25 \leq x~\wedge~x < -\dfrac32~\implies~x \notin \mathbb{R}$