# 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?

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Tex Comment: For future reference, \leq is <= (less than or equal to), \geq is >= (greater than or equal to) and \neq is != (not equal to). –  Eric Naslund Feb 23 '11 at 23:47
See also my answer here. –  Isaac Feb 24 '11 at 1:44
And something should have been shouting in your ear that there was something terribly wrong: your solution set includes $x=-1$, but the inequality is not defined there! –  Arturo Magidin Feb 24 '11 at 3:29

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 $x \in (-1,4]$

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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,

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You deleted your comment? In any case, I edited to add some more details. When we look at case 1 we are assuming $x+1>0$ so that is where the $x>-1$ comes from. –  Eric Naslund Feb 23 '11 at 23:58

To preserve the $\:\le\:$ you must multiply by $\rm\ (x+1)^2\$ not $\rm\ x+1\:,\:$ namely

$\rm\quad\quad\quad\quad\quad\quad\ \displaystyle\frac{2x-3}{x+1}\ \le\ 1$

$\rm\quad\quad\iff\quad \displaystyle\frac{x-4}{x+1}\ \ \le\ 0$

$\rm\quad\quad\iff\quad (x+1)\ (x - 4)\ \le\ 0,\quad x\ne -1$

$\rm\quad\quad\iff\quad\ x\ \in\ (-1,4\:]$

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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.

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