<s>[INST] A question about solving quotients all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:
$$
\begin{eqnarray}
\lvert \frac{x+1}{x-2} \rvert &<& 3\\
\Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac{x+1}{x-2} < 3\\
x+1&>&-3(x-2)\hspace{2cm}\text{(Solving for the left inequality)}\\
x+1&>&-3x+6\\
4x&>&5\\
x&>&\frac{5}{4}
\end{eqnarray}
$$
which cannot be true, as $x$ can equal $2$ with these restrictions which is obviously not allowed. Doing the left side the correct way:
$$
\begin{eqnarray}
\frac{x+1}{x-2} &>& -3\\
\frac{x+1}{x-2} + 3 &>& 0\\
\frac{x+1}{x-2} + \frac{3x-6}{x-2} &>& 0\\
\frac{4x-5}{x-2} &>& 0\\
\Rightarrow x < \frac{5}{4}\lor x>2\\
\end{eqnarray}
$$
leads to the solution set $(-\infty,\frac{5}{4})\cup(2,\infty)$. My question is, what happened with the first method where I failed to come up with a solution for $(2,\infty)$, (and why the signage for $x>\frac{5}{4}$ is backwards/incorrect in the first example, does the multiplication by -3 reverse the inequality even if I'm not introducing a new negative term to one side?). I assume since you cannot divide by zero, you're not allowed to multiply the denominator to the other side without restrictions, but I am not quite sure. Thank you, I greatly appreciate it.
 [/INST] HINT
Since both sides are nonnegative, you can square them in order to obtain an equivalent inequation:
\begin{align*}
\left|\frac{x+1}{x-2}\right| < 3 & \Longleftrightarrow \left(\frac{x+1}{x-2}\right)^{2} < 9\\\\
& \Longleftrightarrow \frac{(x^{2} + 2x + 1) - 9(x^{2} - 4x + 4)}{(x-2)^{2}} < 0\\\\
& \Longleftrightarrow \frac{-8x^{2} + 38x - 35}{(x-2)^{2}} < 0\\\\
& \Longleftrightarrow
\begin{cases}
8x^{2} - 38x + 35 > 0\\\\
x\neq 2
\end{cases}
\end{align*}
Can you take it from here?
</s>