Solve for x: 1/ x + 1 + 3 / 5 x + 1 = 5 / x + 4, x ≠ 1, – 1/5, – 4

Question

Q). Solve for x: $\frac{1}{x + 1} + \frac{3}{5 x + 1} = \frac{5}{x + 4}$, x ≠ 1, - $\frac{1}{5}$, - 4

Sapling Academy · Accepted Answer

$\frac{1}{x + 1} + \frac{3}{5 x + 1} = \frac{5}{x + 4}$

∴ $\frac{1 (5 x + 1) + 3 (x + 1)}{(x + 1)(5 x + 1)} = \frac{5}{x + 4}$

∴ $\frac{5 x + 1 + 3 x + 3}{(x + 1)(5 x + 1)} = \frac{5}{x + 4}$

∴ $\frac{8 x + 4}{(x + 1)(5 x + 1)} = \frac{5}{x + 4}$

∴ (8 x + 4) (x + 4) = 5 (x + 1)(5 x + 1)

∴ 8 x2 + 4 x + 32 x + 16 = 5 ( 5 x2 + 5 x + x + 1)

∴ 8 x2 + 36 x + 16 = 25 x 2 + 25 x + 5 x + 5

∴ 17 x2 - 6 x - 11 = 0

∴ 17 x2 - 17 x + 11 x - 11 = 0         (by mid term splitting)

∴ 17 x (x - 1) + 11 (x - 1) = 0

∴ (x -1) (17 x + 11) = 0

Hence we get x = 1 and x = $\frac{- 11}{17}$

Here we reject x = 1 (because it is given that x ≠ 1); and select x = $\frac{- 11}{17}$

Therefore, value of x = $\frac{- 11}{17}$.

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