F.4續方程 超急

您好！我是 lop，高興能解答您的問題。

(1)

x^2 - 2x - 3y = 0 --- (1)
y + x - 2 = 0 --- (2)

從 (2)

y = 2 - x --- (3)

代 (3) 入 (1)

x^2 - 2x - 3(2-x) = 0
x^2 - 2x - 6 + 3x = 0
x^2 + x - 6 = 0
(x+3)(x-2) = 0
x = -3 或 2 --- (4)

代 x = -3 入 (3)

y = 2 - (-3)
y = 5

代 x = 2 入 (3)

y = 2 - 2
y = 0

答：x = -3 時，y = 5；x = 2 時，y = 0。

(2)

x^2 + y^2 = 5 --- (1)
2x - y = 3 --- (2)

從 (2)

y = 2x - 3 --- (3)

代 (3) 入 (1)

x^2 + (2x-3)^2 = 5
x^2 + 4x^2 - 12x + 9 = 5
5x^2 - 12x + 4 = 0
(x-2)(5x-2) = 0
x = 2/5 或 2

代 x = 2/5 入 (3)

y = 2(2/5) - 3
= 4/5 - 3
= -11/5

代 x = 2 入 (3)

y = 2(2) - 3
= 4 - 3
= 1

答：x = 2/5 時，y = -11/5；x = 2 時，y = 1。