<s>[INST] Solving $|\frac{x+1}{x-2}|<1$ and $\frac{6w+7}{2w-1}- \frac{6w+1}{2w}=1$. I want to find
the real number such that $|\frac{x+1}{x-2}|<1$ and find the values $w$ such that :
$$\frac{6w+7}{2w-1}- \frac{6w+1}{2w}=1$$.
By playing a little bit with this first inequality I have found out that the real numbers $(-\infty,0]$ satisfy the inequality. But when Im trying to prove this I got te following:
$|\frac{x+1}{x-2}|<1$ means that $-1< \frac{x+1}{x-2}<1$ from the left inequality I got $-x+2<x+1$. Then I got $\frac{1}{2}<x$. From the right inequality I got that $x+1 < x-2$, but this gives me no information for $x$ since its cancelled. So what can I conclude from the right inequality? Does all the real numbers satisfy this inequality are $x \in (- \infty,  \frac{1}{2})$?
For the second equality I got that:
$$\frac{6w+7}{2w-1}- \frac{6w+1}{2w}= \frac{6w+7-6w-1}{(2w+1)(2w)}=\frac{6}{4w^{2}+2w}$$
Then I got $0=4w^{2}+2(w)-6$ which can be solved by the quadratic formula but this gives me values $\frac{-6}{4}$ and $1$. But clearly when $w=1$ the equality doesnt hold.
But Im stuck there. Thanks!
 [/INST] Another way to handle the inequality here is by squaring the absolute value symbol away:
$$\begin{align}
\left|x+1\over x-2\right|\lt1
&\iff\left(x+1\over x-2\right)^2\lt1\\
&\iff(x+1)^2\lt(x-2)^2\\
&\iff x^2+2x+1\lt x^2-4x+4\\
&\iff6x\lt3\\
&\iff x\lt1/2
\end{align}$$
so the inequality is satisfied for all $x\in(-\infty,1/2)$.
The one subtlety here is that we need to check that $x=2$, where the original left hand side is undefined, is not in the interval $(-\infty,1/2)$. If, for example, the inequality sign had pointed the other way, the final answer would be $(1/2,2)\cup(2,+\infty)$ instead of $(1/2,+\infty)$.
For the equation in $w$, you made two errors in the first step. It should be
$${6w+7\over2w-1}-{6w+1\over2w}={(6w+7)(2w)-(6w+1)(2w-1)\over(2w-1)(2w)}$$
That is, you forgot to cross multiply in the numerator, and you changed $2w-1$ to $2w+1$ in the denominator.  Can you take things from there?
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