<s>[INST] How do I solve inequalities of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$? I'm stuck trying to solve an inequality of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$; specifically, $\left|\frac{2x + 5}{x + 1}\right| \geq 1$
I tried the approach that was taught for solving inequalities in general, and solved first for x when the part enclosed in absolute value signs is  $\geq 1$, and then solved for when it is $\leq -1$, but plugging in numbers for the resulting domain $[-4, -2]$ back into the original inequality resulted in some answers that were not $\geq 1$.
Where have I gone wrong in trying to solve this inequality? How is this different from solving an inequality like $\left|\frac{3}{x + 1}\right| \geq 1$?
EDIT: Okay, since I'm clearly messing this up horribly(and now realize that I was absent-mindedly taking the intersection and not the union of my results in spite of having not mixed up the two many times before), I'll edit with the specifics of the solution I was trying and which had me confused before I asked the question. So:
$\left|\frac{2x + 5}{x + 1}\right| \geq 1$
Meaning 1: $\frac{2x + 5}{x + 1} \geq 1$ or 2: $\frac{2x + 5}{x + 1} \leq -1$
Tackling the first: multiplied 1 by $x + 1$, forgetting to account for the fact that $x + 1$ could potentially be negative: $2x + 5 \geq x + 1$, then rearrange for $x \geq -4$.
Now the other inequality: multiplied -1 by $x + 1$, again forgetting to consider $x + 1$ being negative: $2x + 5 \leq -x - 1$ , rearrange and divide by 3 for $x \leq -2$.
Domain is the union, so $[-4, \infty] \cup [-\infty, -2]$, so $R - { 1 }$. This is obviously not right, as plugging in -3 shows.
So, in inequalities where the denominator is some function g(x), if I split it and choose to multiply, will I have to solve each part of the inequality twice(once assuming g(x) is positive, a second time assuming it's negative) and then check which solution makes sense by plugging in numbers?
 [/INST] With the restriction that $x \ne -1$,
$\; \left|\dfrac{2x + 5}{x + 1}\right| \geq 1$ is equivalent to
$\dfrac{2x + 5}{x + 1} \ge 1 \qquad$ or $\qquad \dfrac{2x + 5}{x + 1} \le -1$
If $ x + 1 \gt 0$, then
$\quad (2x + 5 \ge x + 1$ and $x > -1)\qquad$ or
$\qquad (2x + 5 \le -x - 1$ and $x > -1)$
$\quad (x \ge -4$ and $x > -1)\qquad$ or
$\qquad (x \le -2$ and $x > -1)$
$\quad x \in (-1, \infty) \cup \varnothing$
If $ x + 1 \lt 0$, then
$\quad (2x + 5 \le x + 1$ and $x < -1)\qquad$ or
$\qquad (2x + 5 \ge -x - 1$ and $x < -1)$
$\quad (x \le -4$ and $x < -1)\qquad$ or
$\qquad (x \ge -2$ and $x < -1)$
$\quad x \in (-\infty, -4] \cup [-2, -1)$

$x \in (-\infty, -4] \cup [-2, -1) \cup(-1, \infty)$
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