Question

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

Answer 1

$\text{Consider,}$

$\displaystyle\frac{a}{ax-1}+\frac{b}{bx-1}=a+b$

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

$\displaystyle\;a(bx-1)+b(ax-1)=(a+b)(ax-1)(bx-1)$

$\displaystyle\;abx-a+abx-b=(a+b)[abx^2-ax-bx+1]$

$\displaystyle\;2abx-(a+b)=(a+b)[abx^2-(a+b)x+1]$

$\displaystyle\;2abx-(a+b)=(a+b)abx^2-(a+b)^2x+(a+b)$

$\displaystyle\;(a+b)abx^2-2abx-(a+b)^2x+2(a+b)=0$

$\displaystyle\;abx((a+b)x-2)-(a+b)((a+b)x-2)=0$

$\displaystyle\;((a+b)x-2)(abx-(a+b))=0$

$\implies\;(a+b)x-2=0\;\text{or}\;abx-(a+b)=0$

$\implies\;x=\frac{2}{a+b},\frac{a+b}{ab}$

$\therefore\textbf{The solution set is}\;\{\bf\frac{2}{a+b},\frac{a+b}{ab}\}$