# Solve for x: 2x-1=(6x^3-5x^2+3x)/(3x^2-x-14)?

Oct 19, 2017

$x = \frac{7}{15}$

#### Explanation:

Given:

$2 x - 1 = \frac{6 {x}^{3} - 5 {x}^{2} + 3 x}{3 {x}^{2} - x - 14}$

Multiply both sides by $\left(3 {x}^{2} - x - 14\right)$ to find:

$6 {x}^{3} - 5 {x}^{2} + 3 x = \left(2 x - 1\right) \left(3 {x}^{2} - x - 14\right)$

$\textcolor{w h i t e}{6 {x}^{3} - 5 {x}^{2} + 3 x} = 6 {x}^{3} - 5 {x}^{2} - 27 x + 14$

Subtract $6 {x}^{3} - 5 {x}^{2} - 27 x$ from both sides to get:

$30 x = 14$

Divide both sides by $30$ to get:

$x = \frac{14}{30} = \frac{7}{15}$