Taking the equation \(5x−1+3y+1=9686\), the last digit of \(5x−1\) will always be 5 for all positive integral values of \(x.\)
The power cycle of 3 is:
\(3^{4k+1}≡3^{4k+2}≡9^{4k+3}≡7^{4k}≡1\)
Clearly, \(3y+1\) must be in the form of \(34k\) as the unit digit of the right-hand side is 6.
We have \(3^4=81\), and \(3^8=6561\). Also, \(9686−81=9605\) and \(9686−6561=3125.\) Observe that \(3125=55.\)
Hence, \(5x−1=55\) or \(x=6\) and \(3y+1=38⇒y=7\) (where \(x=6\) and \(y=7\) also satisfies the first equation).
Therefore, \(x+y=6+7=13.\)