<s>[INST] Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\frac{7}{x+4}\\
-1(x+4)<7\\-x < 11\\
x>-11.$$
This results in the answer $x>-11,  x>3,$ which doesn't make sense.
 [/INST] If $k>0$, then $a<b\iff ak<bk$. However, if $k<0$, then $a<b\iff ak\color{red}{>}bk$.
So when multiplying an inequality by a constant, we should be certain whether that constant is positive or negative. In the case of $-1<\frac{7}{x+4}$, multiplying by $x+4$ is not a good idea because we don't know if $x+4<0$ or $x+4>0$. A better idea is to multiply by $(x+4)^2$, which is positive.* Then,
\begin{align}
-1<\frac{7}{x+4}&\iff-(x+4)^2<7(x+4) \\[5pt]
&\iff -x^2-8x-16<7x+28 \\[5pt]
&\iff 0<x^2+15x+44 \\[5pt]
&\iff0<(x+4)(x+11) \\[5pt]
&\iff x<-11\text{ or }x>-4 \, .
\end{align}

*Technically, $(x+4)^2$ could also be zero, if $x=-4$. However, this is a non-issue, as if $x=-4$, then $7/(x+4)$ is undefined, so it is not a solution to $-1<7/(x+4)$; and $x=-4$ is not a solution to $-(x+4)^2<7(x+4)$ either.
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