Taking 2‌nd ‌ 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: 34k+1≡3 34k+2≡9 34k+3≡7 34k≡1 Clearly 3y+1 must be in the form of 34k as the unit digit of R.H.S.=6 We have 34=81, and 38=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 (x=6 and y=7 also satisfies the first equation) Therefore, x+y=6+7=13