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

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(D) Given the inequality: $\frac{8x^2-14x-9}{3x^2-7x-6} > 2$ \ Subtract $2$ from both sides: $\frac{8x^2-14x-9}{3x^2-7x-6} - 2 > 0$ \ $\frac{8x^2-14

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$(-\infty, -2/3) \cup (3, \infty)$

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(D) Given the inequality: $\frac{8x^2-14x-9}{3x^2-7x-6} > 2$ \ Subtract $2$ from both sides: $\frac{8x^2-14x-9}{3x^2-7x-6} - 2 > 0$ \ $\frac{8x^2-14x-9 - 2(3x^2-7x-6)}{3x^2-7x-6} > 0$ \ $\frac{8x^2-14x-9 - 6x^2+14x+12}{3x^2-7x-6} > 0$ \ $\frac{2x^2+3}{3x^2-7x-6} > 0$ \ Since $2x^2+3$ is always positive for all real $x$,the inequality holds if $3x^2-7x-6 > 0$ \ Factor the denominator: $3x^2-9x+2x-6 = 3x(x-3)+2(x-3) = (3x+2)(x-3) > 0$ \ The critical points are $x = -2/3$ and $x = 3$ \ Testing the intervals $(-\infty, -2/3)$,$(-2/3, 3)$,and $(3, \infty)$,the expression is positive in $(-\infty, -2/3) \cup (3, \infty)$.

The set of all real values of $x$ satisfying the inequation $\frac{8x^2-14x-9}{3x^2-7x-6} > 2$ is

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