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.
 A: In the first method, when you go from $\dfrac{x+1}{x-2}\gt -3$ to $x+1\gt -3(x-2)$, what you're doing is multiplying both sides of the inequality by $x-2$ and you assume that $x-2\gt 0$ when doing that because otherwise the inequality sign flips to $\le$
So, assuming $x\gt 2$, you get from that inequality that $x\gt 5/4$ and the second inequality $\dfrac{x+1}{x-2}\lt 3$ gives $x\gt 7/2$
So, assuming $x\gt 2$, you get $x\gt 5/4$ and $x\gt 7/2$; together they imply $x\gt\max\{2,5/4,7/2\}=7/2$
Similarly, assume $x-2\lt 0$, ie, $x\lt 2$ and solve $x+1\lt -3(x-2)$ and $x+1\gt 3(x-2)$ simultaneously to obtain the other solution set. You get $x\lt\min\{2,5/4,7/2\}=5/4$
The complete solution set is thus $(-\infty,5/4)\cup (7/2,\infty)$

The answer you arrived at is wrong. Note that $x=3\in (2,\infty)$ doesn't satisfy the original equation.
In the second method, you're only solving one of the two inequalities that are supposed to hold. If you solve the second inequality $\dfrac{x+1}{x-2}\lt 3$ in a similar way, you get $$\frac{x+1}{x-2}-3\lt 0\iff\frac{7-2x}{x-2}\lt 0\iff x\in (-\infty, 2)\cup (7/2,\infty)$$
and the solution set to your original problem is the intersection of this with $(-\infty,5/4)\cup (2,\infty)$ so that
$$[(-\infty,2)\cup (7/2,\infty)]\cap[(-\infty,5/4)\cup (2,\infty)]=(-\infty,5/4)\cup (7/2,\infty)$$
which is the solution set to the original problem $\left|\dfrac{x+1}{x-2}\right|\lt 3$
A: 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?
A: We can only "flip the sign" if what is inside the absolute value is negative.  And similarly, we "keep the sign", if what is inside the absolute value bracket is positive.
We could write it this way...
$-3<\frac {x+1}{x-2}<0$ or $0\le \frac {x+1}{x-2}<3$
Or say, the left-hand equation only applies when $-1<x<2$
Then we can solve the two sets of inequalities:
$-1<x<\frac 54$ or  $x > \frac {7}{2}$ or $x \le -1$
Now we can combine the intervals
$(-\infty, \frac 54)\cup (\frac 72,\infty)$
