Question

If x = `(8ab)/(a + b)`, prove that `(x + 4a)/(x - 4a) + (x + 4b)/(x - 4b)` = 2

Answer 1

Given:

x = `(8ab)/(a + b)`

Step 1: Simplify the first fraction using componendo and dividendo

`x/(4a) = (2b)/(a + b)`

Applying componendo and dividendo,

`(x + 4a)/(x - 4a) = (2b + (a + b))/(2b - (a + b))` = `(a + 3b)/(b - a)`

Step 2: Simplify the second fraction using componendo and dividendo

`x/(4b) = (2a)/(a + b)`

Applying componendo and dividendo,

`(x + 4b)/(x - 4b) = (2a + (a + b))/(2a - (a + b)) = (3a + b)/(a - b)`

Step 3: Add the simplified fractions

`(x + 4a)/(x - 4a) + (x + 4b)/(x - 4b) = (a + 3b)/(b - a) + (3a + b)/(a - b)`

Since b − a = −(a − b), we can rewrite the expression with a common denominator:

`(-(a + 3b))/(a - b) + (3a + b)/(a - b)`

= `(-a - 3b + 3a + b)/(a - b)`

= `(2a - 2b)/(a - b)`

= `(2(a - b))/(a - b)`

= 2

∴ It is proven that `(x + 4a)/(x - 4a) + (x + 4b)/(x - 4b)` = 2 when x = `(8ab)/(a + b)`