# How do you combine (3x-2)*[x/(3x^2+x-2)+2/(x+1)]?

Oct 1, 2016

$\left(3 x - 2\right) \cdot \left[\frac{x}{3 {x}^{2} + x - 2} + \frac{2}{x + 1}\right] = \frac{7 x - 4}{x + 1}$

#### Explanation:

Let us first factorize $3 {x}^{2} + x - 2$

$3 {x}^{2} + x - 2 = 3 {x}^{2} + 3 x - 2 x - 2$

= $3 x \left(x + 1\right) - 2 \left(x + 1\right)$

= $\left(3 x - 2\right) \left(x + 1\right)$

Hence $\left(3 x - 2\right) \cdot \left[\frac{x}{3 {x}^{2} + x - 2} + \frac{2}{x + 1}\right]$

= $\left(3 x - 2\right) \cdot \left[\frac{x}{\left(3 x - 2\right) \left(x + 1\right)} + \frac{2}{x + 1}\right]$

= $\left(3 x - 2\right) \times \frac{x}{\left(3 x - 2\right) \left(x + 1\right)} + \left(3 x - 2\right) \times \frac{2}{x + 1}$

= $\cancel{3 x - 2} \times \frac{x}{\cancel{\left(3 x - 2\right)} \left(x + 1\right)} + \frac{2 \left(3 x - 2\right)}{x + 1}$

= $\frac{x}{x + 1} + \frac{2 \left(3 x - 2\right)}{x + 1}$

= $\frac{x + 2 \left(3 x - 2\right)}{x + 1}$

= $\frac{x + 6 x - 4}{x + 1}$

= $\frac{7 x - 4}{x + 1}$