The set of all real values of $x$ satisfying the inequality $\frac{7 x^2-5 x-18}{2 x^2+x-6} \lt 2$ is

Question

Aicharya · Accepted Answer

Since, $\frac{7 x^2-5 x-18}{2 x^2+x-6} \lt 2$ $\Rightarrow \frac{3 x^2-7 x-6}{(x+2)(2 x-3)} \lt 0 \Rightarrow \frac{(x-3)(3 x+2)}{(x+2)(2 x-3)} \lt 0$

So, $x \in\left(-2, \frac{-2}{3}\right) \cup\left(\frac{3}{2}, 3\right)$.

The set of all real values of $x$ satisfying the inequality $\frac{7 x^2-5 x-18}{2 x^2+x-6} \lt 2$ is

Solution