TITLE: How to find the solution for $\frac{2x-3}{x+1} \leq 1$?
QUESTION [8 upvotes]: I have the following inequality:
$$\frac{2x-3}{x+1}\leq1$$
so, considering $x \neq -1$, I started multiplying $x+1$ both sides:
$$2x-3\leq x+1$$
then I subtracted $x$ both sides:
$$x-3\leq1$$
and then sum $3$ both sides:
$$x\leq4$$
Therefore, my solution for $x\neq-1$ is:
$$(-\infty,4]$$
But the book solution is:
$$(-1,4]$$
What I did wrong?

REPLY [7 votes]: To preserve the $\:\le\:$ you must multiply by $\rm\ (x+1)^2\ $ not $\rm\ x+1\:,\:$ namely
$\rm\quad\quad\quad\quad\quad\quad\  \displaystyle\frac{2x-3}{x+1}\ \le\ 1$
$\rm\quad\quad\iff\quad \displaystyle\frac{x-4}{x+1}\ \ \le\ 0 $
$\rm\quad\quad\iff\quad  (x+1)\ (x - 4)\ \le\ 0,\quad x\ne -1 $
$\rm\quad\quad\iff\quad\ x\ \in\ (-1,4\:] $

REPLY [3 votes]: As others have mentioned, multiplying by $x+1$ forces you to consider cases at the outset. Instead, you can write $\frac{2x-3}{x+1} \leq 1$ as $\frac{2x-3}{x+1} -1 \leq 0$. Simplify this into $\frac{p(x)}{q(x)} \leq 0$ and consider when a fraction is negative.