F.4maths續方程

1.
3x + y = 7 ...... [1]
x² + y² =41 ...... [2]

由 [1] :
y = 7 - 3x ...... [3]

把 [3] 代入[2] 中：
x² + (7 - 3x)² =41
x² + 49 - 42x + 9x² =41
5x² - 21x + 4 = 0
(5x - 1)(x - 4) = 0
x = 1/5 或 x = 4

把 x 之值代入[3] 中：

當 x = 1/5：
y = 7 - 3(1/5)
y = 32/5

當 x = 4：
y = 7 - 3(4)
y = -5

答案：x =1/5, y = 32/5 或 x = 4, y = -5


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2.
x² + y² + 6x- 8y = 0 ...... [1]
x + 3y - 14 = 0 ...... [2]

由 [2]：
x = 14 - 3y ...... [3]

把 [3] 代入[1]：
(14 - 3y)² + y² + 6(14- 3y) - 8y = 0
196 - 84y + 9y² + y² +84 - 18y - 8y = 0
y² - 11y + 28 = 0
(y - 4)(y - 7) = 0
y = 4 或 y = 7

把 y 之值代入 [3] 中：

當 y = 4：
x = 14 - 3(4)
x = 2

當 y = 7：
x = 14 - 3(7)
x = -7

答案： x = 2, y = 4 或 x = -7, y = 7


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3.
y + 1 = x + 11 ...... [1]
6x² - 4 = x + 11 ...... [2]

由 [2]：
6x² - x - 15 = 0
(3x - 5)(2x + 3) = 0
x = 5/3 或 x = -3/2

把 x 之值代入 [3] 中：

當 x = 5/3：
y + 1 = (5/3) + 11
y = 35/3

當 x = -3/2：
y + 1 = (-3/2) + 11
y = 17/2

答案： x = 5/3, y = 35/3 或 x = -3/2, y = 17/2


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4.
y = ax² - 4x + 2 ...... [1]
y = 3x + 1 ...... [2]

[1] = [2]：
ax² - 4x + 2 = 3x + 1
ax² - 7x + 1 = 0

由於有兩個交點，故以上方程式有兩個相異的實根，判別式 Δ > 0
(-7)² - 4 * a * 1 > 0
49 - 4a > 0
4a < 49
a < 49/4


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5.
[1/(x + 1)] + [1/(x - 1)] = 3/4
[(x - 1) + (x + 1)] / [(x + 1)(x - 1)] = 3/4
2x / [(x + 1)(x - 1)] = 3/4
3(x + 1)(x - 1) = 4(2x)
3x² - 3 = 8x
3x² - 8x - 3 = 0
(3x + 1)(x - 3) = 0
x = -1/3 或 x = 3

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