How do you solve $\frac{a}{a - 5} + 2 = \frac{3 a}{a + 5}$ $a/(a-5)+2=(3a)/(a+5)$ ?

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How do you solve $\frac{a}{a - 5} + 2 = \frac{3 a}{a + 5}$ $a/(a-5)+2=(3a)/(a+5)$ ?

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Answer 1 · 2015-05-24T13:16:27

Given:

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

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

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

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

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

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

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

$= 20 a - 50$ $=20a-50$

Add $50$ $50$ to both sides to get

$20 a = 50$ $20a=50$

Divide both sides by $50$ $50$ to get:

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

Check by substituting back in the original equation:

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

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

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

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How do you solve $\frac{a}{a - 5} + 2 = \frac{3 a}{a + 5}$ $a/(a-5)+2=(3a)/(a+5)$ ?

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How do you solve $\frac{a}{a - 5} + 2 = \frac{3 a}{a + 5}$ $a/(a-5)+2=(3a)/(a+5)$ ?