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Laws of Exponents for Real Numbers

Evaluate : x...

Question

Evaluate :
$\frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 3$

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Solution

$\frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 3$
$\Rightarrow \frac{(x + 1) (x + 2) + (x - 1) (x - 2)}{(x - 1) (x + 2)} = 3$
$\Rightarrow x^{2} + 2 x + x + 2 + x^{2} - 2 x - x + 2 = 3 (x^{2} + 2 x - x - 2)$
$\Rightarrow 2 x^{2} + 4 = 3 x^{2} + 3 x - 6$
$\Rightarrow x^{2} + 3 x - 10 = 0$
$\Rightarrow x^{2} + 5 x - 2 x - 10 = 0$
$\Rightarrow (x + 5) (x - 2) = 0$
So x =2,-5.

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