# What is the domain of (4/(a+1))-(5/a) = (20/(a^2+a))?

Oct 26, 2015

$a = \left\{- 25\right\}$

#### Explanation:

$\left[1\right] \text{ } \frac{4}{a + 1} - \frac{5}{a} = \frac{20}{{a}^{2} + a}$

Combine $\frac{4}{a + 1}$ and $- \frac{5}{a}$ by making them have the same denominator.

$\left[2\right] \text{ } \frac{4 \left(a\right)}{a \left(a + 1\right)} - \frac{5 \left(a + 1\right)}{a \left(a + 1\right)} = \frac{20}{{a}^{2} + a}$

$\left[3\right] \text{ } \frac{4 \left(a\right) - 5 \left(a + 1\right)}{a \left(a + 1\right)} = \frac{20}{{a}^{2} + a}$

$\left[4\right] \text{ } \frac{4 a - 5 a - 5}{{a}^{2} + a} = \frac{20}{{a}^{2} + a}$

$\left[5\right] \text{ } \frac{- a - 5}{{a}^{2} + a} = \frac{20}{{a}^{2} + a}$

Multiply both sides by $\left({a}^{2} + a\right)$ to remove the denominator.

$\left[6\right] \text{ } \cancel{{a}^{2} + a} \left(\frac{- a - 5}{\cancel{{a}^{2} + a}}\right) = \cancel{{a}^{2} + a} \left(\frac{20}{\cancel{{a}^{2} + a}}\right)$

$\left[7\right] \text{ } - a - 5 = 20$

$\left[8\right] \text{ } - a = 25$

$\left[9\right] \text{ } \textcolor{b l u e}{a = - 25}$