We have,
x−5ax−1=x−bbx−1
(x−5)(bx−1)=(ax−1)(x−b)
bx2−5bx−x+5=ax2−x−abx+b
⇒bx2−5bx+5=ax2−abx+b
⇒bx2−ax2−5bx+abx+5−b=0
⇒x2(b−a)−b(5b−ab)+(5−b)=0
Hence, this is the answer.