Question 476228
{{{5/(y-3) - 30/(y^2-9)=1}}}
(5/y-3)-{30/(y+3)(y-3)}=1    {By Algebraic identity (a^2-b^2)=(a+a)(a-b)}
(5y+15-30)/{(y+3)(y-3)}=1
(5y-15)/{(y+3)(y-3)}=1
{{5(y-3)}/{(y+3)(y-3)}}=1
(5)/(y+3)=1  (By cancelling out y-3 in numerator and denominator)
5=y+3
5-3=y
2=y
y=2