Question

Solve the following inequalities.
$\frac{x^2-7x+12}{x-4}⩾0$

Answer 1

$\frac{x^2-7x+12}{x-4}\ge 0$

Here, $x\ne 4$

Now,

$\frac{(x-4)(x-3)}{x-4}\ge 0$

$⟹x-3\ge 0$

$⟹x\ge 3$ but $x\ne 4$

Hence, solution set is

$x\in [3,\infty )-\left\{4\right\}$