Solve the system

5^(y-x)*(y+x) = 1

and

(x+y)^(x-y) = 5

(In reply to

Puzzle Solution: Method 1 by K Sengupta)

Taking logarithm in base 5, we obtain:

log(5)(x+y) = x-y .......(i)

(x-y)*log(5)(x+y) = 1 ....(ii)

Dividing (i) by (ii), we obtain:

1/(x-y) = x-y, so that:

(x-y)^2 = 1, giving:

x-y = +/-1

If x-y = 1, then log(5)(x+y) = 1, so that: x+y = 5

Solving, we have: (x,y) = (3, 2)

If x-y = -1, then log(5)(x+y) = 11, so that: x+y = 1/5

Solving, we have: (x,y) = (-2/5. 3/5).

Consequently, the required solution is given by (x, y) = (3, 2), or (-2/5, 3/5).