Solve for x ((6x + 2a + 3b + c )/(6x + 2a − 3b − c)) = ((2x + 6a + b + 3c)/(2x + 6a − b − 3c))


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Question Number 14559 by tawa tawa last updated on 02/Jun/17

$Solve for x$ $\frac{6 x + 2 a + 3 b + c}{6 x + 2 a - 3 b - c} = \frac{2 x + 6 a + b + 3 c}{2 x + 6 a - b - 3 c}$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 02/Jun/17

	$\frac{(6 x + 2 a + 3 b + c) + (6 x + 2 a - 3 b - c)}{(6 x + 2 a + 3 b + c) - (6 x + 2 a - 3 b - c)} =$ $= \frac{(2 x + 6 a + b + 3 c) + (2 x + 6 a - b - 3 c)}{(2 x + 6 a + b + 3 c) - (2 x + 6 a - b - 3 c)} \Rightarrow$ $\frac{12 x + 4 a}{6 b + 2 c} = \frac{4 x + 12 a}{2 b + 6 c} \Rightarrow \frac{6 x + 2 a}{3 b + c} = \frac{2 x + 6 a}{b + 3 c}$ $\Rightarrow \frac{6 x + 2 a}{2 x + 6 a} = \frac{3 b + c}{b + 3 c} \Rightarrow \frac{(6 x + 2 a) + 3 (2 x + 6 a)}{(6 x + 2 a) - 3 (2 x + 6 a)} =$ $\frac{(3 b + c) + 3 (b + 3 c)}{(3 b + c) - 3 (b + 3 c)} \Rightarrow \frac{12 x + 20 a}{- 16 a} = \frac{6 b + 10 c}{- 8 c}$ $\Rightarrow 3 x + 5 a = \frac{a}{c} (3 b + 5 c)$ $3 x = \frac{3 a b + 5 a c - 5 a c}{c} = \frac{3 a b}{c} \Rightarrow x = \frac{a b}{c} . ◼$
	Commented by Tinkutara last updated on 02/Jun/17

	$It is known as componendo and$ $dividendo which states that if$ $\frac{a}{b} = \frac{c}{d}, then \frac{a + b}{a - b} = \frac{c + d}{c - d}$ $In next step, \frac{a + 3 b}{a - 3 b} = \frac{c + 3 d}{c - 3 d} is used .$
	Commented by tawa tawa last updated on 02/Jun/17

	$Wow, i really appreciate sir .$
	Commented by tawa tawa last updated on 02/Jun/17

	$God bless you sir .$
	Commented by tawa tawa last updated on 02/Jun/17

	$Sir, i {don}^{'} t really understand the first step . please what did you do sir .$
	Commented by mrW1 last updated on 02/Jun/17

	$i f \frac{a}{b} = \frac{c}{d} t h e n$ $\frac{a + n b}{a - n b} = \frac{c + n d}{c - n d} n ϵ R, n \neq \frac{a}{b}$ $p r o o f :$ $\frac{a + n b}{a - n b} = \frac{\frac{a}{b} + n}{\frac{a}{b} - n} = \frac{\frac{c}{d} + n}{\frac{c}{d} - n} = \frac{c + n d}{c - n d}$
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