Question

Solve for x :
$\sf \dfrac{1}{x - 3} - \dfrac{1}{x + 5} = \dfrac{1}{6}$

Thank You​

Answer 1

Answer:

Hope the answer in the above attachment helps you....

Answer 2

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✏Solve for x :

$\sf \dfrac{1}{x - 3} - \dfrac{1}{x + 5} = \dfrac{1}{6}$

$\huge{ \underline{ \star{ \color{blue}{ \bold{ Answer...}}}}}$

✳ 7 or -9

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⛄GIVEN...

$\sf \dfrac{1}{x - 3} - \dfrac{1}{x + 5} = \dfrac{1}{6}$

⛄TO FIND...

$\large{ \sf{ value \: of \: x...}}$

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$:\large{ \sf{ \implies{ \red{ \frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6} }}}} \\ \\ : \large{ \sf{ \implies{ \red{ \frac{x + 5 - (x - 3)}{(x - 3)(x + 5)} = \frac{1}{6} }}}} \\ \\ : \large{ \sf{ \implies{ \red{ \frac{ \cancel{x} + 5 - \cancel{ x} + 3}{ {x}^{2} + 2x - 15 } = \frac{1}{6} }}}} \\ \\ : \large{ \sf{ \implies{ \red{ \frac{8}{ {x}^{2} + 2x - 15 } = \frac{1}{6} }}}} </p><p>\\{\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \:(cross \: multiplying...)}} \\ \\ : \large{ \sf{ \implies{ \red{ {x}^{2} + 2x - 15 = 48}}}} \\ \\ : \large{ \sf{ \implies{ \red{ {x}^{2} + 2x - 63 = 0}}}} \\ \\: \large{ \sf{ \implies{ \red{ {x}^{2} + 9x - 7x - 63 = 0}}}} \\ \\ : \large{ \sf{ \implies{ \red{x(x + 9) - 7(x + 9) = 0}}}} \\ \\ : \large{ \sf{ \implies{ \red{(x - 7)(x + 9) = 0}}}} \\ \\ : \large{ \sf{ \implies{ \boxed{ \green{x = 7 \: or \: - 9}}}}}$

$\large{ \bold{ \purple{ \underline{required \: answer \: is \: 7 \: or \: - 9}}}}$

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✒In order to factorise x2+ ax+ b, we need to find numbers p and q such that p+q= a and pq= b

✒After spilting, we need to take common factors and then we will get something like (x+p)(x+q)