Question

Solve the following equation by factorization

${\displaystyle \frac{a}{ax-1}+\frac{b}{bx-1}=a+b,a+b\ne 0,ab\ne 0}$

Answer 1

${\displaystyle \frac{a}{ax-1}+\frac{b}{bx-1}=a+b,a+b\ne 0,ab\ne 0}$

⇒ ${\displaystyle (\frac{a}{ax-1}-b)+(\frac{b}{bx-1}-a)}$ = 0

⇒ ${\displaystyle \frac{a-abx+b}{(ax-1)}+\frac{b-abx+a}{(bx-1)}}$ = 0

⇒ ${\displaystyle (a+b-abx)[\frac{1}{ax-1}+\frac{1}{bx-1}]}$ = 0

⇒ ${\displaystyle (a+b-abx)\left[\frac{bx-1+ax-1}{(ax-1)(bx-1)}\right]}$ = 0

⇒ ${\displaystyle \frac{(a+b-abx)(ax+bx-2)}{(ax-1)(bx-1)}}$ = 0

⇒ ${\displaystyle (a+b-abx)(ax+bx-2)}$ = 0
⇒ Either a + b - abx = 0,
then a+ b = abx

x = ${\displaystyle \frac{a+b}{ab}}$
or
ax + bx - 2 = 0,
then x(a + b) = 2

x = ${\displaystyle \frac{2}{a+b}}$

Hence x = ${\displaystyle \frac{a+b}{ab},\frac{2}{a+b}}$.