The set of all real values of x satisfying th | vedclass

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(B) Given the inequality: $\frac{7 x^2-5 x-18}{2 x^2+x-6} < 2$Subtract $2$ from both sides: $\frac{7 x^2-5 x-18}{2 x^2+x-6} - 2 < 0$$\Rightarrow \frac

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$\left(-2, -\frac{2}{3}\right) \cup \left(\frac{3}{2}, 3 \right)$

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(B) Given the inequality: $\frac{7 x^2-5 x-18}{2 x^2+x-6} Subtract $2$ from both sides: $\frac{7 x^2-5 x-18}{2 x^2+x-6} - 2 $\Rightarrow \frac{7 x^2-5 x-18 - 2(2 x^2+x-6)}{2 x^2+x-6} $\Rightarrow \frac{7 x^2-5 x-18 - 4 x^2-2 x+12}{(2 x-3)(x+2)} $\Rightarrow \frac{3 x^2-7 x-6}{(2 x-3)(x+2)} Factor the numerator: $3 x^2-7 x-6 = 3 x^2-9 x+2 x-6 = 3 x(x-3)+2(x-3) = (3 x+2)(x-3)$So,the inequality becomes: $\frac{(3 x+2)(x-3)}{(2 x-3)(x+2)} The critical points are $x = -2, -\frac{2}{3}, \frac{3}{2}, 3$.Using the wavy curve method,we test the intervals:For $x > 3$,the expression is positive.For $\frac{3}{2} For $-\frac{2}{3} For $-2 For $x Since we need the expression to be less than $0$,the solution is $x \in \left(-2, -\frac{2}{3}\right) \cup \left(\frac{3}{2}, 3 \right)$.

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

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