# How do you solve (x^2-x-6)/(x+2)+(x^3+x^2)/x=3?

Jan 17, 2017

$x = - 1 \pm \sqrt{7}$

#### Explanation:

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

$\textcolor{w h i t e}{3} = \frac{\left(x - 3\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}}} + \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \left({x}^{2} + x\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}$

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

Subtract $3$ from both ends to get:

$0 = {x}^{2} + 2 x - 6$

$\textcolor{w h i t e}{0} = {x}^{2} + 2 x + 1 - 7$

$\textcolor{w h i t e}{0} = {\left(x + 1\right)}^{2} - {\left(\sqrt{7}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x + 1\right) - \sqrt{7}\right) \left(\left(x + 1\right) + \sqrt{7}\right)$

$\textcolor{w h i t e}{0} = \left(x + 1 - \sqrt{7}\right) \left(x + 1 + \sqrt{7}\right)$

Hence:

$x = - 1 \pm \sqrt{7}$

These are both solutions of the original equation, since neither value causes any denominator to be $0$.