Question

$\frac{x + 3}{x - 3} + \frac{x - 3}{x + 3} = 2 \times \frac{1}{2}$
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Answer 1

Step-by-step explanation:

$\frac{(x + 3) {}^{2} + (x - 3) {}^{2} }{(x {}^{2} - 9)} = 2 \times \frac{1}{2} \\ \frac{x {}^{2} + 9 + 6x + x {}^{2} + 9 - 6x }{x {}^{2} - 9 } \\ \frac{2x {}^{2} + 18}{x {}^{2} - 9} = 1 \\ 2x {}^{2} + 18 = x {}^{2} - 9 \\ 2x {}^{2} - x {}^{2} = - 9 - 18 \\ x {}^{2} = - 27$

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Answer 2

Step-by-step explanation:

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

$\frac{(x + 3)}{(x - 3)} + \frac{(x - 3)}{(x + 3)} = 1$

Taking LCM

$\frac{(x + 3)(x + 3) + (x - 3)(x - 3)}{(x - 3)(x + 3)} = 1$

$\frac{( {x}^{2} + 6x + 9) + ( {x}^{2} - 6x + 9) }{(x - 3)(x + 3)} = 1$

$\frac{ {x}^{2} + 6x + 9 + {x}^{2} - 6x + 9 }{(x - 3)(x + 3)} = 1$

$\frac{ {2x}^{2} + 18}{ {x}^{2} - {3}^{2} } = 1$

by cross multiplying

2x²+18 = x² -9

2x²-x²= -18-9

x² = -27

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