Question

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

Answer 1

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

⇒`[a/((ax-1))-b]+[b/((bx-1))-a]=0`  

⇒`(a-b(ax-1))/(ax-1)+(b-a(bx-1))/(bx-1)=0` 

⇒`(a-abx+b)/(ax-1+a)-(abx+b)/(bx-1)=0` 

⇒`(a-abx+b)[1/(ax-1)+1/(bx-1)]=0` 

⇒`(a-abx+b)[((bx-1)+(ax-1))/(((ax-1))(bx-1))]=0` 

⇒`(a-abx+b)[[(a+b)x-2)/((ax-1)(bx-1))]=0` 

⇒`(a-abx+b)[(a+b)x-2]=0` 

⇒`a-abx+b=0  or  (a+b)x-2=0` 

⇒`x=(a+b)/(ab)  or  x=2/(a+b)` 

Hence, the roots of the equation are `(a+b)/(ab)`  and` 2/(a+b)`