Question

12 question please solve

Answer 1

your solution is here

step by step explaintion
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Answer 2

Question:

If $\dfrac{3x}{(x-6)(x-a)} = \dfrac{2}{x-6} + \dfrac{1}{x+a}$, then a =

Solution:

$\dfrac{3x}{(x-6)(x-a)} = \dfrac{3}{x-6} + \dfrac{1}{x+a}$

$\dfrac{3x}{(x-6)(x-a)} = \dfrac{2(x+a) + (x-6)}{(x-6)(x+a)}$

$\dfrac{3x}{\cancel{(x-6)(x-a)}} = \dfrac{2(x+a) + (x-6)}{\cancel{(x-6)(x+a)}}$

$3x = 2(x+a) + (x-6) \\ 3x = 2x + 2a + x - 6 \\ 3x = 3x + 2a - 6 \\ 3x - 3x = 2a - 6 \\ 0 = 2a - 6 \\ a = \dfrac{6}{2} = 3$

Therefore, a = 3