solving for Inequalities with has modulus function on both numerator and denominator

Question

solve $$\frac{|x|-1}{|x|-3} \ge 0$$

here $x$ is not equal to $3$ and $-3$.

Answer 1

Well a quotient is non negative if the numerator is zero, or if the numerator and the denominator have the same sign.

So case 1 : $|x|-1 \geq 0$ and $|x|-3 > 0$. This is equivalent to $x \in (-\infty, -1] \cup [1, +\infty)$ and $x \in (-\infty, -3) \cup (3, +\infty)$, i.e. you get that $$ x \in (-\infty, -3) \cup (3, +\infty)$$

Case 2 : $|x|-1 \leq 0$ and $|x|-3 < 0$. This is equivalent to $x \in [-1,1]$ and $x \in (-3,3)$, so you get $$x \in [-1,1]$$

Finally, the solutions $S$ are the union of these two cases, i.e. $$S = (-\infty, -3) \cup [-1,1] \cup (3, +\infty)$$

Answer 2

Hint : as ${|x|-1\over|x|-3} \ge 0 \implies $ numerator and denominator are either BOTH POSITIVE or BOTH NEGATIVE $\implies|x|-1\ge0 $ and $|x|-3\gt0$

• or

$\implies|x|-1\le0 $ and $|x|-3\lt0$

Answer 3

Multiply both sides by the denominator $|x| - 3$, but you must split into two cases depending on whether it is positive or negative.

If it is positive, then you get $|x| -1 \geq 0$, whence $|x| \geq 1$. The assumption that the denominator was positive is equivalent to $|x| > 3$, and so, combining these two requirements, we get a solution here iff $|x| > 3$.

If it is negative, then you get $|x| -1 \leq 0$, whence $|x| \leq 1$, and the assumption that the denominator was negative is equivalent to $|x| < 3$, so combining these two requirements, we get a solution in this case iff $|x| \leq 1$.

Then the set of solutions is $\{x \, : \, |x| \leq 1\} \cup \{x \, : \, |x| > 3\} = \left(-\infty, -3\right) \cup [-1,1] \cup \left(3, \infty\right)$