maths(解方程 二 : 代數方法)

15. 設兩數為x及y
x + y = 1 ... (1)
1/y - 1/x = 3/2 ... (2)
(1) => y = 1 - x
代入(2), 1 / (1 - x) - 1/x = 3/2
(x - 1 + x) / [x(1 - x)] = 3/2
2(2x - 1) = 3x(1 - x)
4x - 2 = 3x - 3x^2
3x^2 + x - 2 = 0
(3x - 2)(x + 1) = 0
x = 2/3 或 x = -1
x = 2/3 => y = 1/3較大的一個是2/3
x = -1 => y = 2較大的一個是2
30. (a) 分母較分子大3 => q = p + 3 ... (1)
(p + 2) / (q + 2) = 2(p/q)
(p + 2) / (q + 2) = 2p/q
q(p + 2) = 2p(q + 2)
pq + 2q = 2pq + 4p
2q = pq + 4p ... (2)
(b) 代(1)入(2), 2(p + 3) = p(p + 3) + 4p
2p + 6 = p^2 + 3p + 4p
p^2 + 5p - 6 = 0
(p + 6)(p - 1) = 0
p = -6(不合)或p = 1
p = 1 => q = 4
該分數為1/4
32. 設較長的路線速度為 x km/h
設較短的路線速度為 y km/h
20 / y - 24 / x = 0.5 (0.5 hour = 30 min) ... (1)
x - y = 4
y = x - 4 ... (2)
代(2)入(1), 20 / (x - 4) - 24 / x = 0.5
20x - 24x + 96 = 0.5x(x - 4)
96 - 4x = 0.5x^2 - 2x
0.5x^2 + 2x - 96 = 0
x^2 + 4x - 192 = 0
(x + 16)(x - 12) = 0
x = 12 或 x = -16 (不合)
他騎腳踏車走較長的路線需時= 24/12 = 2小時
33. 錦松在20歲時存入的 $1000到22歲本利和為1000(1 + r/100)^2
在21歲時存入的 $1900到22歲本利和為1900(1 + r/100)
1000(1 + r/100)^2 + 1900(1 + r/100) = 3300
1000(1 + 2r/100 + r^2/10000) + 1900 + 19r = 3300
1000 + 20r + r^/10 + 1900 + 19r = 3300
r^2/10 + 39r - 400 = 0
r^2 + 390r - 4000 = 0
(r - 10)(r + 400) = 0
r = 10或r = - 400(不合)