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If ๐‘ฅ =6โข๐‘Žโข๐‘๐‘Ž+๐‘, find the value of ๐‘ฅ+3โข๐‘Ž๐‘ฅโˆ’3โข๐‘Ž +๐‘ฅ+3โข๐‘๐‘ฅโˆ’3โข๐‘. - Mathematics

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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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