maths

Answer: y = 1 and x = -1

Solution:

x² + 2xy + 2y² - 2y + 1 = 0
x² + 2xy + y² + y² - 2y + 1 = 0
[(x)² + 2(x)(y) + (y)²] + [(y)² - 2(y)(1) + (1)²] = 0
(x + y)² + (y - 1)² = 0

since (x + y)² ≥ 0 and (y - 1)² ≥ 0 and (x + y)² + (y - 1)² = 0,
a unique solution are x + y = 0 and y - 1 = 0
it follows y = 1 and x = -1

只係簡單的因式分解 無需2個等式吧@@