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.
Best Answer
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 $\left|\dfrac{x+1}{x-2}\right|\lt 3$