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Question
If `x = (6ab)/(a + b)`, find the value of `(x + 3a)/(x - 3a) + (x + 3b)/(x - 3b)`.
Sum
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Solution
x = `(6ab)/(a + b)`
⇒ `x/(3a) = (2b)/(a + b)`
Applying compinendo and dividend,
`(x + 3a)/(x - 3a) = (2b + a + b)/(2b - a - b)`
`(x + 3a)/(x - 3a) = (3b + a)/(b - a)` ...(1)
Again, `x = (6ab)/(a + b)`
⇒ `x/(3b) = (2a)/(a + b)`
Applying componendo and dividendo,
`(x + 3b)/(x - 3b) = (2a + a + b)/(2a - a - b)`
`(x + 3b)/(x - 3b) = (3a + b)/(a - b)` ...(2)
From (1) and (2)
`(x + 3a)/(x - 3a) + (x + 3b)/(x - 3b) = (3b + a)/(b -a) + (3a + b)/(a - b)`
= `(-3b -a + 3a + b)/(a - b)`
= `(2a - 2b)/(a - b)`
= `(2(a-b))/(a-b)`
= 2
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