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

Feb 17, 2017

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

#### Explanation:

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

$6 > \left(x + 1\right) \left(3 x - 4\right)$

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

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

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

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

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

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

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

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