Simultaneous equations F.2

6y - (x-4)/3 = 10 ... (1)
x + (3y-7)/4 = 1 ...(2)
(1) => 18y - (x - 4) = 30
18y - x = 26 ... (3)
(2) + (3) => (3y - 7)/4 + 18y = 1 + 26
3y - 7 + 72y = 108
75y = 115
y = 23/15
代入(3), 18(23/15) - x = 26
27.6 - x = 26
x = 1.6
設每個洋娃娃的售價是x
Let the selling price of each doll be x

6y-(x-4)/3=10 ...(1)
x+(3y-7)/4=1 ...(2)

From (1),
6y=10+(x-4)/3
6y=(x+26)/3
y=(x+26)/18 ...(3)

Put (3) into (2),
x+[3(x+26)/18-7]/4=1
x+[(x-16)/6]/4=1
x+(x-16)/24=1
24x+x-16=24
25x=40
x=8/5*

Put x=8/5 into (3),
y=(8/5+26)/18
=23/15**

*8/5=1.6
**23/15=1.53(cor. to 3 sig. fig.)

設每個洋娃娃的售價是x
注意:設題既時候需要寫單位
所以這句話是:
Let $x be the (selling) price of each*** doll
selling可以寫可以唔寫,視乎你計算既題目有冇其他意思相似既字(e.g.原價,現價)
each可以用a或者one代替

2009-12-30 17:26:06 補充：
樓上Let字後面用is一定錯!!
Let後面要用bare infinitive!!

let the value of toy is $x,

6y-(x-4)/3=10 ----(i)

x+(3y-7)/4=1-----(ii)

(i)x3 :18y-x+4=30----(iii)

(iii)+ (ii): 18y+4+(3y-7)/4=30+1

y=75/22

sub y=75/22 into (iii)

x=389/11

6y-(x-4)/3=10...(1)
x+(3y-7)/4=1...(2)

(1)*3; (2)*4
18y-x+4=30...(3)
4x+3y-7=4...(4)

From (3) x=18y-26. Sub. into (4)
4(18y-26)+3y-11=0
y=23/15,x=1.6

設每個洋娃娃的售價是x
Let the selling price of each doll is x

6y-(x-4)/3=10----(1)
x+(3y-7)/4=1---(2)

from (1),
6y-(x-4)/3=10
18y-x+4=30
72y-4x+16=120
72y-4x=104---(3)

from (2),
x+(3y-7)/4=1
4x+3y-7=4
4x+3y=11---(4)

(3)+(4)
75y=115
y=115/75
y=23/15

from (4)
4x+3*23/15=11
4x=11-23/5
4x=32/5
x=8/5


設每個洋娃娃的售價是x的英文 = let x be the price of each doll