Question

solve this inequality

Answer 1

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Answer 2

Step-by-step explanation:

We have,

$\sf \dfrac{2x+3}{x^2+x-12} \leq \dfrac{1}{2}$

$\sf \dfrac{2x+3}{x^2+x-12} - \dfrac{1}{2}\leq 0$

$\sf \dfrac{4x+6-x^2-x+12}{2(x^2+x-12)} \leq 0$

$\sf \dfrac{-x^2+3x+18}{x^2+x-12} \leq 0$

$\sf \dfrac{x^2-3x-18}{x^2+x-12} \geq 0$

$\sf \dfrac{(x-6)(x+3)}{(x-3)(x+4)} \geq 0$

$\sf x \in ( - \infty ,-4 \big] \cup (-3, 3\big] \cup (6, \infty )$