# How do you solve a/(a-5)+2=(3a)/(a+5)?

May 24, 2015

Given:

$\frac{a}{a - 5} + 2 = \frac{3 a}{a + 5}$

Multiply through by $\left({a}^{2} - 25\right) = \left(a - 5\right) \left(a + 5\right)$ to get:

$a \left(a + 5\right) + 2 \left({a}^{2} - 25\right) = 3 a \left(a - 5\right) = 3 {a}^{2} - 15 a$

Subtract $3 {a}^{2} - 15 a$ from both sides to get:

$0 = a \left(a + 5\right) + 2 \left({a}^{2} - 25\right) - 3 {a}^{2} + 15 a$

$= {a}^{2} + 5 a + 2 {a}^{2} - 50 - 3 {a}^{2} + 15 a$

$= \left({a}^{2} + 2 {a}^{2} - 3 {a}^{2}\right) + \left(5 a + 15 a\right) - 50$

$= 20 a - 50$

Add $50$ to both sides to get

$20 a = 50$

Divide both sides by $50$ to get:

$a = \frac{5}{2}$

Check by substituting back in the original equation:

LHS $= \frac{a}{a - 5} + 2 = \frac{\frac{5}{2}}{\frac{5}{2} - 5} + 2$

$= \frac{5}{5 - 10} + 2 = \frac{5}{-} 5 + 2 = - 1 + 2 = 1$

RHS $= \frac{3 a}{a + 5} = \frac{3 \left(\frac{5}{2}\right)}{\frac{5}{2} + 5} = \frac{3 \left(\frac{5}{2}\right)}{3 \left(\frac{5}{2}\right)} = 1$