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Mathematics

If x=6aba+bx = \dfrac{6ab}{a + b}, prove that (x+3ax−3a+x+3bx−3b)=2\Big(\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}\Big) = 2.

Ratio Proportion

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Answer

Given,

x=6aba+bx=3a×2ba+bx3a=2ba+bx = \dfrac{6ab}{a + b} \\[1em] x = \dfrac{3a \times 2b}{a + b} \\[1em] \dfrac{x}{3a} = \dfrac{2b}{a + b}

Applying componendo and dividendo rule,

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

Again solving x,

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

Apply the Componendo and Dividendo rule,

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

Adding equations (1) and (2), we get :

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

Hence, proved that (x+3ax−3a+x+3bx−3b)=2\Big(\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}\Big) = 2.

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