I'm thinking:
S1:
3x + 2y = 12
(3/2)x + y = 6
Since y and 6 are both integers, (3/2)x must be an integer too. That means x is a multiple of 2, so it must be 2 or 4.
If x = 2, then y = 3, and we're fine.
If x = 4, then y = 0, but that's forbidden, since y is positive.
If x > 4, then y < 0, which is also forbidden, so our only solution is x = 2, y = 4.
S2:
|9x² + 30xy + 25y²| = 21²
or
9x² + 30xy + 25y² = ±21²
(3x + 5y)² = ±21²
Since the left side is positive, the right side must be positive:
(3x + 5y)² = 21²
So (3x + 5y) = 21 or (3x + 5y) = -21. But the second case is impossible, since x and y are both positive.
That leaves 3x + 5y = 21. If y = 1, 2, or 4, x is not an integer. If y > 4, then x is negative. We're left with y = 3, forcing x = 2, the only solution.
