solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$. 
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.

options
$a.)\ -101<x<25\\
b.)\ [-\infty,3]\\
c.)\ x\leq 3\\
\color{green}{d.)\ x<3}\\
$
I tried ,
Case $1$ ,for $ \boxed{x\geq 0}\\
\dfrac{x^2-x-12}{x-3}\geq 2x\\
\implies \dfrac{x^2-5x+12}{x-3}\leq 0 \\
\implies x<3\\
x\in \emptyset $
Case $2$ ,for $\boxed{x< 0}\\
\dfrac{x^2+x-12}{x-3}\geq 2x\\
\implies \dfrac{(x-4)(x-3)}{x-3}\leq 0 \\
\implies x\leq 4\\
\implies x< 0\\
  $
But the answer given is option $d.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.

The answer is (d), which is the union of the solutions for the two cases.
