Question

Solve the following equation by factorisation method:

`a/(ax-1) + b/(bx-1) = a+b, a+b \cancel= 0, ab \cancel= 0`

Answer 1

Given:

`a/(ax-1) + b/(bx-1) = a+b, a+b \cancel= 0, ab \cancel= 0`

LCM = (ax − 1) (bx − 1)

`(a(bx-1) + b(ax-1))/((ax-1)(bx-1)) = a+b`

a(bx − 1) + b(ax − 1)

= abx − a + abx − b

= 2abx − (a + b)

`(2abx - (a+b))/((ax-1)(bx-1)) = a+b`

2abx − (a + b) = (a + b) (ax − 1) (bx − 1)

(ax − 1) (bx − 1) = abx² − (a + b)x + 1

⇒ (a + b) [abx² − (a + b)x + 1]

2abx − (a + b) = (a + b) [abx² − (a + b)x + 1]

(a + b) [abx² − (a + b)x + 1] − 2abx + (a + b) = 0

ab(a + b)x² − [(a + b)² + 2ab]x + 2(a + b) = 0

(abx − (a + b) (a + b)x − 2) = 0

abx − (a + b) = 0

⇒ x = `(a+b)/(ab)`

(a + b)x − 2 = 0

⇒ x = `2/(a+b)`

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