How many ordered pairs of positive integers (x, y) satisfy 4x + 7y = 141?

x = (141 − 7y)/4 must be a positive integer.

141 − 7y ≡ 0 (mod 4) → 7y ≡ 141 ≡ 1 (mod 4) → 3y ≡ 1 (mod 4) → y ≡ 3 (mod 4) (since 3 × 3 = 9 ≡ 1).

Also 7y < 141 → y ≤ 20. So y ∈ {3, 7, 11, 15, 19}.

Corresponding x: 30, 23, 16, 9, 2 — all positive.

There are 5 pairs.

Tip: Once one solution (30, 3) is found, the others are obtained by x → x − 7, y → y + 4 (the coefficients swapped).