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Mathematics

If x = 8aba+b,\dfrac{8ab}{a + b}, find the value of

x+4ax−4a+x+4bx−4b.\dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b}.

Ratio Proportion

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Answer

Given,

x=8aba+bx4a=8aba+b4a∴x4a=2ba+bx = \dfrac{8ab}{a + b} \\[1em] \dfrac{x}{4a} = \dfrac{\dfrac{8ab}{a + b}}{4a} \\[1em] \therefore \dfrac{x}{4a} = \dfrac{2b}{a + b}

By componendo and dividendo, ⇒x+4ax−4a=2b+a+b2b−a−b⇒x+4ax−4a=3b+ab−a[….Eq 1]x4b=8aba+b4b∴x4b=2aa+b\Rightarrow \dfrac{x + 4a}{x - 4a} = \dfrac{2b + a + b}{2b - a - b } \\[0.5em] \Rightarrow \dfrac{x + 4a}{x - 4a} = \dfrac{3b + a}{b - a} \qquad \text{[….Eq 1]} \\[1.5em] \dfrac{x}{4b} = \dfrac{\dfrac{8ab}{a + b}}{4b} \\[1em] \therefore \dfrac{x}{4b} = \dfrac{2a}{a + b}

By componendo and dividendo,

⇒x+4bx−4b=2a+a+b2a−a−b⇒x+4bx−4b=3a+ba−b[….Eq 2]\Rightarrow \dfrac{x + 4b}{x - 4b} = \dfrac{2a + a + b}{2a - a - b} \\[0.5em] \Rightarrow \dfrac{x + 4b}{x - 4b} = \dfrac{3a + b}{a - b} \qquad \text{[….Eq 2]}

Adding Eq 1 and 2,

⇒x+4ax−4a+x+4bx−4b=3b+ab−a+3a+ba−b⇒x+4ax−4a+x+4bx−4b=3b+ab−a−3a+bb−a⇒x+4ax−4a+x+4bx−4b=3b−b+a−3ab−a⇒x+4ax−4a+x+4bx−4b=2b−2ab−a⇒x+4ax−4a+x+4bx−4b=2(b−a)b−a=2\Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{3b + a}{b - a} + \dfrac{3a + b}{a - b} \\[1em] \Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{3b + a}{b - a} - \dfrac{3a + b}{b - a} \\[1em] \Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{3b - b + a - 3a}{b - a} \\[1em] \Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{2b - 2a}{b - a} \\[1em] \Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{2(b - a)}{b - a} = 2 \\[1em]

Hence, the required value is 2.

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