Question

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

Answer 1

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

Step 1: Simplify the first term using the componendo and dividendo

⇒ `x/a = (2b)/(a + b)`

Apply Componendo and Dividendo,

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

Step 2: Simplify the second term using the componendo and dividendo

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

Apply Componendo and Dividendo,

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

Step 3: Combine the simplified terms

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

We can rewrite the second denominator, as a − b = −(b − a)

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

Now, combine the numerators over the common denominator:

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

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

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

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

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

= 2

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