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Mathematics

If x = 6aba+b\dfrac{6ab}{a + b}, find the value of :

x+3ax−3a+x+3bx−3b\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}.

Ratio Proportion

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Answer

(i) Given,

⇒x=6aba+b⇒x3a=2ba+b\Rightarrow x = \dfrac{6ab}{a + b} \\[1em] \Rightarrow \dfrac{x}{3a} = \dfrac{2b}{a + b}

Applying componendo and dividendo:

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

Again,

⇒x=6aba+b⇒x3b=2aa+b\Rightarrow x = \dfrac{6ab}{a + b} \\[1em] \Rightarrow \dfrac{x}{3b} = \dfrac{2a}{a + b}

Applying componendo and dividendo:

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

Adding (i) and (ii) we get,

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

Hence, x+3ax−3a+x+3bx−3b\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b} = 2.

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