How to find the solution for $\frac{2x-3}{x+1} \leq 1$? 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?
 A: 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\:] $
A: The problem is when you multiplied both sides by $x+1$.  Remember that when you multiply an inequality by a negative number it changes signs.  This means we have to split into cases:
Case 1:  $x+1>0$.  Then we get $$2x-3\leq x+1$$
By the same reasoning that you present above we find $x\leq 4$.  Then rewrite the inequality $x+1>0$ as $x>-1$.  Combining this with $x\leq 4$ we see $x\in (-1,4]$.
Case 2: $x+1<0$.  Then we get $$2x-3\geq x+1$$ (notice that since $x+1<0$ the sign had to switch directions)  We then solve to find $x\geq 4$.  Since for this case we also had $x+1<0$, which is the same as $x<-1$ we conclude no such $x$ exists.  (A number cannot be less than -1 and greater than 4)
Hope that helps,
A: Note that if we have $\frac{a}{b} \leq 1$, this means that if $b>0$, then $a \leq b$ and if $b<0$, then $a \geq b$.
So you will need to split this into cases.
First note that you can multiply without changing the $\leq$ sign only when $x+1 > 0$.
So when $x+1 > 0$, we have $2x-3 \leq x+1$ which gives us $x \leq 4$. Hence, when $x+1>0$, we have $x \leq 4$ and hence $x \in (-1,4]$.
If $x+1 < 0$, the $\leq$ gets reversed to $\geq$ when you multiply by $x+1$ and we get $2x-3 \geq x+1$ when $x+1 < 0$. This gives us $x \geq 4$ when $x < -1$ which is not possible.
Hence, the solution is $x \in (-1,4]$
A: 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.
