數學一問用解方程

1.長子所得的 = x ;
次子所得的 = 1/4 x – 400

長子所得的 +次子所得的 = 6000
x + (1/4 x – 400) = 6000
5/4 x – 400 = 6000
5/4 x = 6400
5x = 25600
x = 5120

長子所得的 = 5120
次子所得的 = 1/4 (5120) – 400
= 1280 -400
= 880

2. 小份的= x
大份的 = 45 – x

5 (45 – x ) – 1 = 3x
225 – 5x – 1 = 3x
224 = 8x
x = 28

小份的= 28
大份的 = 45 – 28
= 17


3. 五角硬幣數目的 = x
一元硬幣的數目 = 2x-2
二角硬幣的數目 = (2x - 2) + 6
= 2x + 4

0.5x + (2x - 2) + 0.2 (2x +4) = 27.8
0.5x + 2x - 2 + 0.4x + 0.8 = 27.8
2.9 x – 1.2 = 27.8
2.9 x = 29
x = 10

五角硬幣數目的 = 10

一元硬幣的數目 = 2x-2
= 2 (10) – 2
= 20 - 2
= 18

4.
A有的錢 = x
B有的錢 = x – 60

(x – 9 ) = 3 ((x – 60) + 9)
x – 9 = 3x – 153
153 – 9 = 3x – x
144 = 2x
x = 72

A原有的錢 =72

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你既方法雖然都係有同樣的結果
但邏輯上講唔通
一個十分鮮明既講法對比:
假設花為白色的 並唔等於 假設白色的為花

2.) 設x為次子所得的金額.
設y為長子所得的金額.
x+y=6000 --- (1)
y/4=x+400 --- (2)
由(2),
y=4x+1600 --- (3)
代(3)入(1),
x+(4x+1600)=6000
5x = 4400
x = 880

3.) 設x為小份的朱古力數.
設y為大份的朱古力數.
x+y=45 --- (1)
5x-1=3y --- (2)
由(1),
x=45-y --- (3)
代(3)入(2),
5(45-y)-1=3y
225-5y-1=3y
224=8y
28=y
y=28
所以 x=45-y
= 27

4.) 設x為一元硬幣數目.
設y為五角硬幣數目.
設z為二角硬幣數目.
x+2=2y --- (1)
z-6=x --- (2)
x+0.5y+0.2z=27.8 --- (3)
由(1),
y=(x+2)/2 --- (4)
由(2),
z=x+6 --- (5)
代(4), (5) 入 (6),
x+0.5*(x+2)*0.5+0.2(x+6)=27.8
20x+5(x+2)+4(x+6)=556
20x+5x+10+4x+24=556
29x=522
x=18

5.)設x為A原有的金錢.
設y為B原有的金錢.
x-y=60 --- (1)
(x-9)=3(y+9) --- (2)
由(1),
x=60+y --- (3)
代(3)入(2),
(60+y-9)=3(y+9)
51+y=3y+27
24 = 2y
y = 12
所以, x=60+12
x=72

2. 設次子所得的金額是y
y+4(y+400) = 6000
y+4y+1600 = 6000
5y = 4400
y = 880

3. 設小份有y顆巧克力
y+(5y-1)/3 = 45
3y+(5y-1) = 135
8y = 136
y = 17

4. 設一元硬幣的數目是y
y+0.5x[(y+2)/2]+0.2x(y+6) = 27.8
2y+0.5x(y+2)+0.2x2x(y+6) = 55.6
2y+0.5y+1+0.4y+2.4 = 55.6
2.9y = 52.2
y = 18

設A原有y元
y-60 = (y-9)/3
3y-180 = y-9
2y = 171
y = 85.5