二元一次應用題

設M是男士的數目,F是女士的數目
[1/2 (M)+1/10 (F)]x 1.1=253
M/2+F/10=230-----------------(1)
1/4 (M)+1/2 (F)=250
M/4+F/2=250-------------------(2)
∴M=400,F=300
∴舞會的賓客共有700名

2006-11-04 15:50:41 補充：
sorry,我跳了step(1): M/2+F/10=230(2)x2:M/2+F=500------(3)(3)-(1)=>9F/10=270F=300put F=300 into (3),we haveM/2+300=500M/2=200M=400∴M=400,F=300

設共有ｍ名男士及ｆ名女士會出席舞會.
((1/2)m + (1/10)f) * (1+10%) = 253.......(1)
(1/4)m + (1/2)f = 250........(2)

由(2),
(1/4)m + (1/2)f = 250
m + 2f = 1000
m = 1000 - 2f...........(3)

把(3)代入(1),
((1/2)(1000 - 2f) + (1/10)f) * (1+10%) = 253
((500 - f) + (1/10)f) * (110/100) = 253
((500 - f) + (1/10)f) = 253 * (100/110)
((500 - f) + (1/10)f) = 2530/11
500 - f + 0.1f = 2530/11
5500 - 11f + 1.1f = 2530
5500 - 2530 = 9.9f
2970/9.9 = f
f = 300

把f = 300代入(3)
m = 1000 - 2*300
m = 1000 - 600
m = 400

所以共有300女士及400男士, 共700人出席舞會.

設男士有a名
設女士有b名
設出席舞會的賓客共有(a + b)名

Given (1/2a + 1/10b) * (110%) = 253
(1/2a + 1/10b) * 110 = 253 * 100
55a + 11b = 25300 ……(i)
Given 1/4a + 1/2b = 250
a + 2b = 1000
a = 1000 - 2b
Put (a = 1000 - 2b) into (i), we get
55(1000 - 2b) + 11b = 25300
55000 - 110b + 11b = 25300
55000 - 25300 = 110b - 11b
29700 = 99b
300 = b
a = 300
Thus, a = 1000 - 2b = 1000 - 2 * 300 = 1000 - 600 = 400

Thus, a + b = 300 + 400 = 700 (名)

Ans. 出席舞會的賓客共有700名.