How do you solve $\frac{6}{3 x - 4} > x + 1$ $6/(3x-4) > x + 1$ ?

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How do you solve $\frac{6}{3 x - 4} > x + 1$ $6/(3x-4) > x + 1$ ?

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Answer 1 · 2017-02-17T18:48:27

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

Explanation:

Multiply both sides of the inequality by $3 x - 4$ $3x-4$ :

$6 > (x + 1) (3 x - 4)$ $6 > (x+1)(3x-4)$

Distribute the right-side: $6 > (3 x^{2} - 4 x + 3 x - 4)$ $6 > (3x^2-4x+3x-4)$

Simplify by adding like terms and subtract the $6$ $6$ from both sides:
$6 - 6 > 3 x^{2} - x - 4 - 6$ $6-6 > 3x^2-x-4-6$ ;
$0 > 3 x^{2} - x - 10$ $0 > 3x^2-x-10$

Factor the right-side: $0 > (3 x + 5) (x - 2)$ $0 > (3x+5)(x-2)$ or $(3 x + 5) (x - 2) < 0$ $(3x+5)(x-2) < 0$

Solve for values of $x$ $x$ that make the equation negative:
$3 x + 5 < 0$ $3x+5 < 0$ and $x - 2 < 0$ $x-2<0$

$3 x < - 5$ $3x < -5$ and $x < 2$ $x < 2$

$x < - \frac{5}{3}$ $x < -5/3$

Therefore: $- \frac{5}{3} < x < 2$ $-5/3 < x < 2$

From the graph of $(3 x + 5) (x - 2) < 0$ $(3x+5)(x-2) < 0$ you can see when $y < 0$ $y < 0$ :
graph{3x^2-x-10 [-20.97, 19.03, -12.32, 7.68]}

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How do you solve $\frac{6}{3 x - 4} > x + 1$ $6/(3x-4) > x + 1$ ?

1 Answer

Explanation:

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