Solve:
$\frac{1}{x + 2} + \frac{1}{x + 4} = \frac{3}{4}$

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Solution

$\frac{1}{x + 2} + \frac{1}{4} = \frac{3}{4} \frac{x + 4 + x + 2}{x^{2} + 6 x + 8} = \frac{3}{4} 8 x + 24 = 3 x^{2} + 18 x + 24 3 x^{2} + 10 x = 0 x = 0; \frac{- 10}{3}$

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Similar questions

\frac{3}{2} l o g_{1 / 4} (x + 2)^{2} - 3 = l o g_{1 / 4} (4 - x)^{3} + l o g_{1 / 4} (x + 6)^{3}

4 (1 - x) + 3 (2 - x) = - 4

4/x+2 - 1/x+3 = 4/2x+1

Solve using factorization method.

\frac{x - 1}{3} - 1 = \frac{x - 2}{4}

\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}; x = \neq 2, 4

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