Find the range of the values of x

Question

$\frac{(-3)}{2x-7} \leqslant x$

This my attempt:

The answer from me: $ \frac{1}{2} \leqslant x , x \geqslant 3 $

However the correct answer are :$ \frac{1}{2} \leqslant x \leqslant 3 $ or $x \geqslant \frac{7}{2} $

Answer 1

Hint:

We need $$0\ge\dfrac{-3}{2x-7}-x=-\dfrac{3-7x+2x^2}{2x-7}=-\dfrac{(2x-1)(x-3)}{2x-7}$$

Check for the following cases:

$2x-1<=>0$

$2x-7<=>0$

$x-3<=>0$

Answer 2

Rewrite as

$$0\le \frac3{2x-7}+x=\frac{3+2x^2-7x}{2x-7}=\frac{(x-3)(x-\frac12)}{x-\frac72}.$$

The RHS is positive when there is an even number of negative factors.

$$\begin{matrix}&\frac12&&3&&\frac72\\\hline3&&2&&1&&0\end{matrix}$$