Question

solve (1/x-1)-(1/x+5)=6/7

Answer 1

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Answer 2

$\bold{\underline{Hey\: mate \: !}}$

$\bold{\large{\underline{Answer\: \colon}}}$

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$\bold{\large{1). \:\frac{1}{x-1}-\frac{1}{x+5}\: = \: \frac{6}{7}}}$

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$\bold{\large{\underline{Take\: LCM \: \colon}}}$

$\bold{\large{\frac{(x+5)\: -\:(x-1)}{(x-1)(x+5)}\: =\: \frac{6}{7} }}$

$\bold{\large{\frac{x+5-x+1}{x^{2}+5x-x-5}\: = \: \frac{6}{7} }}$

$\bold{\large{\frac{5+1}{x^{2} + 4x -5} \: = \: \frac{6}{7} }}$

$\bold{\large{\frac{6}{x^{2} + 4x -5}\: = \: \frac{6}{7}}}$

$\bold{\underline{Cross \: multiplication\: \colon}}$

$\bold{\large{7\times \frac{6}{6} = x^{2} + 4x -5}}$

$\bold{\large{ 7 = x^{2} + 4x - 5}}$

$\bold{\underline{\large{Change\: places\: sign\: would\: remain \:same\: \colon}}}$

$\bold{\large{ x^{2} + 4x - 5 = 7 }}$

$\bold{\large{ x^{2} + 4x - (5 + 7) = 0}}$

$\bold{\large{ x^{2} + 4x - (12) = 0}}$

$\bold{\large{ x^{2} + 6x - 2x - 12 = 0}}$

$\bold{\large{ x ( x \: + \: 6) \: - 2( x \: + \: 6) \: = \: 0 }}$

$\bold{\large{ ( x \: + \: 6) ( x\: - \: 2) \: = \: 0}}$

Solutions for x :

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$\bold{\large{\boxed{ x \: = \: 2 \: , \: -6}}}$

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$\bold{\large{Thanks}}$