# How do you solve 1/v+(3v+12)/(v^2-5v)=(7v-56)/(v^2-5v) and check for extraneous solutions?

Sep 15, 2016

$v = 21$

#### Explanation:

$\frac{1}{v} + \frac{3 v + 12}{{v}^{2} - 5 v} = \frac{7 v - 56}{{v}^{2} - 5 v}$

$\frac{1}{v} + \frac{3 v + 12}{{v}^{2} - 5 v} - \frac{7 v - 56}{{v}^{2} - 5 v} = 0$

The common denominator is ${v}^{2} - 5 v = v \left(v - 5\right)$

$\frac{v - 5 + 3 v + 12 - \left(7 v - 56\right)}{{v}^{2} - 5 v} = 0$

$\frac{v - 5 + 3 v + 12 - 7 v + 56}{{v}^{2} - 5 v} = 0$

$\frac{v + 3 v - 7 v - 5 + 12 + 56}{{v}^{2} - 5 v} = 0$

$\frac{- 3 v + 63}{{v}^{2} - 5 v} = 0$

$- 3 v + 63 = 0$
$- 3 v = - 63$
$v = \frac{- 63}{- 3}$
$v = 21$