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

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Answer 1 · 2017-02-17T18:48:27

$-5/3 < x < 2$

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

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

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

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

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

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

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

$x < -5/3$

Therefore: $-5/3 < x < 2$

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

How do you solve 6/(3x-4) > x + 16/(3x-4) > x + 1?