How do I solve the inequality $\frac{2x-3}{x+1}\leq 1$?
Here is my thought process:
$2x-3 \leq x+1$
so
$x \leq 4$
So in interval notation, this is $(-\infty,4]$ but apparently this is not correct. What am I doing wrong?
$\endgroup$33 Answers
$\begingroup$$$\dfrac{2x-3}{x+1}\le1\iff0\ge\dfrac{2x-3}{x+1}-1=\dfrac{x-4}{x+1}$$
If $\dfrac{x-4}{x+1}=0, x-4=0$
Else $\dfrac{x-4}{x+1}<0\iff(x-4)(x+1)<0$
Now $(x-a)(x-b)<0$ with $a<b\implies a<x<b$
$\endgroup$4$\begingroup$Hint: what happens with inequalities if you multiply with a negative number? Are you sure that you always multiply with something positive?
$\endgroup$2$\begingroup$$\dfrac{2x - 3}{x + 1} \le 1$
$\iff \dfrac{2x - 3}{x + 1} - 1 \le 0$
$\iff \dfrac{2x - 3}{x + 1} - \dfrac{x + 1}{x + 1} \le 0$
$\iff \dfrac{x - 4}{x + 1} \le 0$
Notice that $\dfrac{x - 4}{x + 1}$ changes sign at $x = -1$ and $x = 4$, so the possible solutions are $(-\infty, -1)$, $(-1, 4]$ and $[4, \infty)$. Take a test value from each of the three candidate solutions to substitute into the inequality. You will observe that $(-1, 4]$ is the only interval that satisfies the inequality.
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