# How do you solve \frac{ 3}{ x } + \frac{ x }{ x + 2} = \frac{ - 2}{ x + 2}?

May 17, 2017

$x = - 3$

#### Explanation:

Since both sides of the equation have $x + 2$ as a common denominator, we can subtract $\frac{x}{x + 2}$ from both sides

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

$\frac{3}{x} + \cancel{\frac{x}{x + 2} \textcolor{red}{- \frac{x}{x + 2}}} = \frac{- 2}{x + 2} \textcolor{red}{- \frac{x}{x + 2}}$

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

Multiply both sides by $- 1$

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

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

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

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

$- \frac{3}{x} = 1$

Multiply both sides by $x$

$- \frac{3}{\cancel{x}} \cancel{\textcolor{red}{\left(x\right)}} = 1 \textcolor{red}{\left(x\right)}$

$- 3 = x$

$x = - 3$

May 17, 2017

color(purple)(x=-3

#### Explanation:

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

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

multiply L.H.S AND R.H.S. by color(purple)( x(x+2)

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

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

$\therefore 5 x + {x}^{2} = - 6$

$\therefore {x}^{2} + 5 x + 6 = 0$

$\therefore \left(x + 3\right) \left(x + 2\right) = 0$

$\therefore x + 3 = 0$ or $x + 2 = 0$

:.color(purple)(x=-3 or color(purple)(x=-2

substitute color(purple)(x=-3

$\therefore \frac{3}{\textcolor{p u r p \le}{- 3}} + \frac{\textcolor{p u r p \le}{- 3}}{\textcolor{p u r p \le}{- 3} + 2} = \frac{- 2}{\textcolor{p u r p \le}{- 3} + 2}$

$\therefore - 1 + \frac{- 3}{-} 1 = \frac{- 2}{- 1}$

$\therefore - 1 + 3 = 2$

$\therefore 2 = 2$

substitute color(purple)(x=-2

$\therefore \frac{3}{\textcolor{p u r p \le}{- 2}} + \frac{\textcolor{p u r p \le}{- 2}}{\textcolor{p u r p \le}{- 2} + 2} = \frac{- 2}{\textcolor{p u r p \le}{- 2} + 2}$

$\therefore - 1.5 + \frac{- 2}{0} = \frac{- 2}{0}$

extraneous solution, so $x$ can't be $- 2$