Question

Solve the following equation by factorisation method:

`1/(2a+b+2x) = 1/(2a) + 1/b + 1/(2x), x \cancel= 0, a \cancel= 0, b \cancel= 0, x \cancel= -((2a+b))/2`

Answer 1

Given:

`1/(2a+b+2x) = 1/(2a) + 1/b + 1/(2x), x \cancel= 0, a \cancel= 0, b \cancel= 0, x \cancel= -((2a+b))/2`

LCM of 2a, b, 2x = 2abx

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

2abx = (2a + b + 2x) (bx + 2ax + ab)

First factor common terms:

bx + 2ax + ab = x(b + 2a) + ab

2abx = (2x + 2a + b) [x(2a + b) + ab]

Expand RHS:

= (2x + 2a + b) [(2a + b)x + ab]

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

= 2(2a + b)x² + 2abx + (2a + b)2x + ab(2a + b)

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

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

(2a + b) [2x² + (2a + b)x + ab] = 0

Since 2a + b ≠ 0,

2x² + (2a + b)x + ab = 0

2x² + (2a + b)x + ab = (2x + b) (x + a)

2x + b = 0

⇒ x = −`b/2`

x + a = 0

⇒ x = −a

x = −a or x = −`b/2`