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Mathematics

If x = 2aba+b\dfrac{2ab}{a + b}, find the value of : x+ax−a+x+bx−b\dfrac{x + a}{x - a} + \dfrac{x + b}{x - b}.

Ratio Proportion

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Answer

Given,

⇒x=2aba+b⇒xa=2ba+b\Rightarrow x = \dfrac{2ab}{a + b} \\[1em] \Rightarrow \dfrac{x}{a} = \dfrac{2b}{a + b}

Applying componendo and dividendo:

⇒x+ax−a=2b+a+b2b−(a+b)⇒x+ax−a=3b+ab−a ........(i)\Rightarrow \dfrac{x + a}{x - a} = \dfrac{2b + a + b}{2b - (a + b)} \\[1em] \Rightarrow \dfrac{x + a}{x - a} = \dfrac{3b + a}{b - a} \space ……..(i)

Also,

⇒x=2aba+b⇒xb=2aa+b\Rightarrow x = \dfrac{2ab}{a + b} \\[1em] \Rightarrow \dfrac{x}{b} = \dfrac{2a}{a + b}

Applying componendo and dividendo:

⇒x+bx−b=2a+a+b2a−(a+b)⇒x+bx−b=3a+ba−b ......(ii)\Rightarrow \dfrac{x + b}{x - b} = \dfrac{2a + a + b}{2a - (a + b)} \\[1em] \Rightarrow \dfrac{x + b}{x - b} = \dfrac{3a + b}{a - b} \space ……(ii)

Adding (i) and (ii) we get,

⇒x+ax−a+x+bx−b=3b+ab−a+3a+ba−b=3b+ab−a+(−3a+bb−a)=3b+ab−a−3a+bb−a=3b−3a+a−bb−a=2b−2ab−a=2(b−a)(b−a)=2.\Rightarrow \dfrac{x + a}{x - a} + \dfrac{x + b}{x - b} = \dfrac{3b + a}{b - a} + \dfrac{3a + b}{a - b} \\[1em] = \dfrac{3b + a}{b - a} + \Big(-\dfrac{3a + b}{b - a}\Big) \\[1em] = \dfrac{3b + a}{b - a} - \dfrac{3a + b}{b - a} \\[1em] = \dfrac{3b - 3a + a - b}{b - a} \\[1em] = \dfrac{2b - 2a}{b - a} \\[1em] = \dfrac{2(b - a)}{(b - a)} \\[1em] = 2.

Hence, x+ax−a+x+bx−b\dfrac{x + a}{x - a} + \dfrac{x + b}{x - b} = 2.

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