Решим систему через формулу разности квадратов a^2 - b^2 = (a - b)(a + b).
(3 + x)^2 - (x + 2)^2 = 4y = [(x+3) - (x+2)] · [(x+3) + (x+2)] = 1 · (2x + 5) = 4y => 2x + 5 = 4y => y = (2x + 5)/4 = (1/2)x + 5/4.
(2 - y)^2 - (1 + y)^2 = -12x = [(2 - y) - (1 + y)] · [(2 - y) + (1 + y)] = (1 - 2y) · 3 = 3 - 6y => 3 - 6y = -12x => y = 2x + 1/2.
Сопоставим выражения для y: (1/2)x + 5/4 = 2x + 1/2 => 5/4 - 1/2 = 2x - (1/2)x => 3/4 = (3/2)x => x = 1/2.
Тогда y = 2x + 1/2 = 2·(1/2) + 1/2 = 1 + 1/2 = 3/2.
Ответ: x = 1/2, y = 3/2.