<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] Alternative
Make sure you only multiply with positive number because otherwise the inequality sign will reverse
$$
\begin{align}
\left|\frac{7}{x+4}\right|\cdot\left|\frac{7}{x+4}\right|&<1\cdot\left|\frac{7}{x+4}\right|\\
\\
&<1\cdot 1\\
\\
\\
7^{2}&<(x+4)^{2}
\end{align}
$$
From here we get the solutions:
$$
x<-11\text{ or }x>3
$$
Notice that I only multiply with $\left|\frac{7}{x+4}\right|$ and $(x+4)^{2}$ which I know are positive numbers.
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