Understanding absolute values with inequalities

Question

I have trouble understanding the following solution of this inequality:

$ \frac{4|x|-2}{1-|x|} ≥ 1$

I move the right side to the left and multiply to get rid of the fraction

$4|x|-2 - 1 (1 - |x|) ≥ 0\\5|x| - 3 ≥ 0\\case1: x ≤ \frac{-3}{5}\\case2: x ≥ \frac{3}{5}$

Now the Solution to this question is $-1 < x ≤ \frac{-3}{5} \\ and \\ \frac{3}{5} ≤ x < 1$

How do I get the ones from the solution? Is it because I cannot divide by zero and by interpreting the equation I set restrictions that it cannot equal to zero?

Answer 1

• A fraction $\frac ab$ is greater ( or equal) than $1$ in 2 possible and mutually exclusive cases :

(1) $a$ and $b$ are both negative and $|a|\geq |b|$

(2) $a$ and $b$ are both positive and and $a \geq b$ ( $a, b \neq 0$)

• Could we be in the first case? It would first require both the numerator and the denominator to be negative.

$4|x| - 2 \lt 0 \iff 4|x| \lt 2\iff |x|\lt \frac 12 \iff x\in (-\frac 12 ; \frac 12)$

$1-|x| \lt 0 \iff - |x| \lt -1 \iff |x| \gt 1 \iff x\lt -1 \lor x\gt 1$

We see that the first case requires inconsistent conditions.

• So we are in the second case, in which both $a$ and $b$ are positive, and $a\geq b$ ( and $a, b \neq 0$).

We therefore need :

$4|x| -2 \gt 0 \iff |x|\gt \frac 12 \iff (x \lt - \frac 12 \lor x\gt \frac 12)$

and

$1-|x| \gt 0 \iff |x| \lt 1 \iff ( x\gt -1 \land x\lt 1)$

meaning that either $x\in (-1; -\frac 12)$ OR $x\in (\frac 12; 1) $

• Keeping in mind these conditions, we can solve the simple inequation $ a\geq b$ :

$4|x| - 2 \geq 1 - |x|$

$\iff 5|x| -3 \geq 0$

$ \iff |x| \geq \frac 35$

$ \iff (x \geq \frac 35) \lor (-x \geq \frac 35)$

$ \iff (x \geq \frac 35) \lor (x \leq \ -\frac 35)$

• Combining our conditions , namely that : either $x\in (-1; -\frac 12)$ OR $x\in (\frac 12; 1) $ , and our solution :

$$ (x \geq \frac 35) \lor (x \leq \ -\frac 35)$$

we finally get :

$x\in (-1 ; - \frac 35]$ OR $ x\in [\frac 35; 1)$.

Answer 2

It is incorrect to multiply both sides by $1-|x|$ unless you include the condition that $1-|x|>0$.

You must consider the other cases separately. For $1-|x|<0$, the inequality will reverse.

Answer 3

Without cases:

It's $$\frac{4|x|-2}{1-|x|}-1\geq0$$ or $$\frac{4|x|-2-(1-|x|)}{1-|x|}\geq0$$ or $$\frac{5|x|-3}{1-|x|}\geq0$$ or $$\frac{5|x|-3}{|x|-1}\leq0$$ or $$\frac{3}{5}\leq|x|<1,$$ which by the definition of the absolute value gives $$\left(-1,-\frac{3}{5}\right]\cup\left[\frac{3}{5},1\right)$$