聯立二元一次方程

1.Let x be the price of a lunch box
Let y be the price of a glass of orange juice

2x+y=46......(1)
3x+2y=74....(2)

(1)*2=
4x+2y=92....(3)

(3)-(2)=
x=18
Sub. x=18 into (1)
2(18)+y=46
y=10

Ans. The price of a lunch box is

2.Let p be the hundred's digit of the original no.
Let q be the unit's digit of the original no.

p+q=12...(1)
100p+50+q-(100q+50+p)=594
99p-99q=594
p-q=6......(2)

(1)+(2)=
2p=18
p=9

Sub. p=9 into (2)
9-q=6
q=3

Ans. The original number is 953

good!

a)
設1個飯盒的價格是x元,
1瓶橙汁的價格是y元

2x+y=46 -(1)

3x+2y=74 -(2)

從 (1),

y=46-2x -(3)

把 (3) 代入 (2)

3x+2(46-2x)=74

3x+92-4x=74

-x=-18

x=18

把x=18代入 (3)

y=46-2(18)

y=10

所以,解是x=18,y=10

b)

設百位數是x,
個位數則是y

x+y=12 -(1)

(x*100+50+y)-(y*100+50+x)=594 -(2)

從(1)

x=12-y -(3)

把(3)代入(2)

[100(12-y)+50+y]-[100y+50+12-y]=594

1200-100y+50+y-100y-50-12+y=594

-198y+1198=594

-594=-198y

y=3

把y=3代入(3)

x=12-3

x=9

所以,解是x=9,y=3

原先的數字是953

Question1`
Let $x and $ y be the price of a lunch box and a glass of orange juice respectively.
2x+y=46......(1)
3x+2y=74......(2)
2x(2)-(1)x3,
2(3x+2y)-3(2x+y)=72x2-46x3
6x+4y-6x-3y=144-138
y=6
Sub.y=6 into (1)
2x+(6)=46
2x=40
x=20
∴ The price of a lunch box is $20 and a glass of orange juice is $6.
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Question2`

Let x and y be the hundred's digit of the original number and the unit's digit of the original number repectively.

p+q=12...(1)
p-q=6......(2)

From(1), 100p+50+q-(100q+50+p)=594
99p-99q=594

(1)+(2),
2p=18
p=9

Sub. p=9 into (2)
9-q=6
q=3

∴ The number is 953.

2008-02-21 20:18:18 補充：
Let $x and $ y be the price of a lunch box and a glass of orange juice respectively.2x+y=46......(1)3x+2y=74......(2)2x(2)-(1)x3,2(3x+2y)-3(2x+y)=74x2-46x36x+4y-6x-3y=10 y=10

2008-02-21 20:18:38 補充：
Sub.y=10 into (1)2x+(10)=462x=36x=18∴ The price of a lunch box is $18 and a glass of orange juice is $10.sorry ar-.-