# How do you solve (x+2)/(x+3)-(x^2)/(x^2-9)=1-((x-1)/(3-x))?

Nov 22, 2017

$x = \frac{- 3 - \sqrt{57}}{4}$ or $\frac{- 3 + \sqrt{57}}{4}$

#### Explanation:

$\frac{x + 2}{x + 3} - \frac{{x}^{2}}{{x}^{2} - 9} = 1 - \left(\frac{x - 1}{3 - x}\right)$

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

Observe that we cannot have $x = \pm 3$. Now multiplying each term by $\left(x + 3\right) \left(x - 3\right)$, we get

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

or ${x}^{2} - x - 6 - {x}^{2} = {x}^{2} - 9 + {x}^{2} + 2 x - 3$

or $- x - 6 = 2 {x}^{2} + 2 x - 12$

or $2 {x}^{2} + 3 x - 6 = 0$

or $x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \cdot 2 \cdot \left(- 6\right)}}{2 \cdot 2}$

= $\frac{- 3 \pm \sqrt{9 + 48}}{4}$

i.e. $x = \frac{- 3 - \sqrt{57}}{4}$ or $\frac{- 3 + \sqrt{57}}{4}$