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Solving an Equation

If x = 8aba...

Question

If $x = \frac{8 a b}{a + b}$ , then find the value of $\frac{x + 4 a}{x - 4 a} + \frac{x + 4 b}{x - 4 b}$

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Solution

We substitute $x = \frac{8 a b}{a + b}$ in the expression $\frac{x + 4 a}{x - 4 a} + \frac{x + 4 b}{x - 4 b}$ as shown below:

$\frac{x + 4 a}{x - 4 a} + \frac{x + 4 b}{x - 4 b} = \frac{\frac{8 a b}{a + b} + 4 a}{\frac{8 a b}{a + b} - 4 a} + \frac{\frac{8 a b}{a + b} + 4 b}{\frac{8 a b}{a + b} - 4 b} = \frac{\frac{8 a b + 4 a (a + b)}{a + b}}{\frac{8 a b - 4 a (a + b)}{a + b}} + \frac{\frac{8 a b + 4 b (a + b)}{a + b}}{\frac{8 a b - 4 b (a + b)}{a + b}}$
$= (\frac{8 a b + 4 a (a + b)}{a + b} \times \frac{a + b}{8 a b - 4 a (a + b)}) + (\frac{8 a b + 4 b (a + b)}{a + b} \times \frac{a + b}{8 a b - 4 b (a + b)}) = \frac{8 a b + 4 a (a + b)}{8 a b - 4 a (a + b)} + \frac{8 a b + 4 b (a + b)}{8 a b - 4 b (a + b)} = \frac{4 a (2 b + a + b)}{4 a (2 b - a - b)} + \frac{4 b (2 a + a + b)}{4 b (2 a - a - b)} = \frac{2 b + a + b}{2 b - a - b} + \frac{2 a + a + b}{2 a - a - b}$
$= \frac{a + 3 b}{- a + b} + \frac{3 a + b}{a - b} = - \frac{(a + 3 b)}{a - b} + \frac{3 a + b}{a - b} = \frac{- a - 3 b + 3 a + b}{a - b} = \frac{2 a - 2 b}{a - b}$
$= \frac{2 (a - b)}{a - b} = 2$

Hence, $\frac{x + 4 a}{x - 4 a} + \frac{x + 4 b}{x - 4 b} = 2$

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