The values of x for which the inequality frac | vedclass

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(D) Given inequality: $\frac{8x^2+16x-51}{(2x-3)(x+4)} > 3$Subtract $3$ from both sides: $\frac{8x^2+16x-51 - 3(2x^2+5x-12)}{(2x-3)(x+4)} > 0$Simplify

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$x  \frac{5}{2}$ or $-3 < x < \frac{3}{2}$

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(D) Given inequality: $\frac{8x^2+16x-51}{(2x-3)(x+4)} > 3$Subtract $3$ from both sides: $\frac{8x^2+16x-51 - 3(2x^2+5x-12)}{(2x-3)(x+4)} > 0$Simplify the numerator: $8x^2+16x-51 - 6x^2-15x+36 = 2x^2+x-15$Factor the numerator: $2x^2+6x-5x-15 = 2x(x+3)-5(x+3) = (2x-5)(x+3)$So,the inequality becomes: $\frac{(2x-5)(x+3)}{(2x-3)(x+4)} > 0$The critical points are $x = -4, -3, \frac{3}{2}, \frac{5}{2}$.Using the wavy curve method (sign scheme) for the intervals $(-\infty, -4), (-4, -3), (-3, \frac{3}{2}), (\frac{3}{2}, \frac{5}{2}), (\frac{5}{2}, \infty)$:The expression is positive in the intervals $(-\infty, -4) \cup (-3, \frac{3}{2}) \cup (\frac{5}{2}, \infty)$.

The values of $x$ for which the inequality $\frac{8x^2+16x-51}{(2x-3)(x+4)} > 3$ holds,are

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