Question

$\frac{a}{(x - a)} + \frac{b}{(x - c)} = \frac{2c}{(x - c)}$
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Answer 1

Answer:

$x \: = \dfrac{a(b - c)}{(a + 2b - 2c)}$

Step-by-step explanation:

Given--->

$\dfrac{a}{(x - a)} + \dfrac{b}{(x - c)} = \dfrac{2c}{(x - c)}$

To find ---->

$value \: of \: x$

Solution---->

$\dfrac{a}{(x - a)} + \dfrac{b}{(x - c)} = \dfrac{2c}{(x - c)}$

$= > \dfrac{a}{(x - a)} + \dfrac{b}{(x - c)} = \dfrac{c}{(x - c)} + \dfrac{c}{(x - c)}$

$= > \dfrac{a}{(x - a)} - \dfrac{c}{(x - c)} = \dfrac{c}{(x - c)} - \dfrac{b}{(x - c)}$

$= > \dfrac{a(x - c) - c(x - a)}{(x - a) \: (x - c)} = \dfrac{(c - b)}{(x - c)}$

$(x - c) \: cancel \: out \: from \: each \: side \: we \: get$

$= > \dfrac{ax - ac - cx + ac}{(x - a) } = (c - b)$

$= > \dfrac{ax - cx}{(x - a)} = (c - b)$

$= > (a - c)x \: = (c - b) \: (x - a)$

$= > (a - c)x \: = (c - b)x - a(c - b)$

$= > (a - c)x - (c - b)x = - a(c - b)$

$= > (a - c - c + b)x = - a(c - b)$

$= > (a + b - 2c)x \: = a(b - c) \\ = > \: \: \: x \: \: \: = \dfrac{a(b - c)}{(a + b - 2c)}$

Answer 2

$\huge{\blue{Answer}}$

Given,

a/(x-a) + b/(x-c) = 2c/(x-c)

To find the value of x ,

Solution:-

a/(x-a) + b/(x-c) = 2c/(x-c)

⇒ a/(x-a) + b/(x-c) = c/(x-c) + c/(x-c)

⇒ a/(x-a) - c/(x-c) = c/(x-c) - b/(x-c)

⇒ a(x-c) - c(x-a) / (x-a) (x-c) = (c-b)/(x-c)

(x-c) cancel out from each side , we get,

⇒ ax - ac - cx + ac / (x-a) = (c-b)

⇒ ax - cx / (x-a) = (c-b)

⇒ (a-c)x = (c-b) (x-a)

⇒ (a-c)x = (c-b)x - a(c-b)

⇒ (a-c)x - (c-b)x = -a(c-b)

⇒ (a-c-c+b)x = -a(c-b)

⇒ (a+b-2c)x = a(b-c)

⇒ x = a(b-c) / (a+b-2c)