<s>[INST] Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$ 
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stuck here.
 [/INST] $$ \frac{4}{x+4} < 2 $$
(We'll do a case-by-case analysis here, which I feel is slightly more illuminating.)
Case 1: $x+4 > 0 \quad$ ($x > -4$)
In this case, we just multiply both sides by $x+4$ to get:
$$ 4 \le 2(x+4) $$
$$ 4 \le 2x + 8 $$
$$ -4 \le 2x $$
$$ -2 \le x $$
So this solution occurs when both $-4 < x$ and $-2 \le x$, which is the same condition as $x \ge -2$. (Draw a number line to see why this is the case.)
Case 2: $x+4 < 0 \quad$ ($x < -4$)
Now, we must flip the inequality when we multiply both sides.
$$ 4 \le 2(x+4) $$
$$ 4 \le 2x + 4 $$
$$ 0 \le 2x $$
$$ 0 \le x $$
$$ x \ge 0 $$
We have a solution whenever $x < -4$ and $x \ge 0$, in other words, $x < -4$. 
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