[中二maths] 聯立二元一次方程-應用題

Let x be the no. of pentagon, y be the no of hexagon
(to make a pentagon, one will uses 5 sticks and hexagon 6)

therefore:

5x+6y=100 ..........(1) and
6(2y)+5(x-6)=100 ...........(2) (i.e. dobule hexagon no and pentagon no reduced by 6)

(1)
6y=100-5x ..........(3)

put (3) into (2)
2(100-5x)+5(x-6)=100
200-10x+50-30=100
170-5x-100=0
70=5x
x=14

Therefore orignal no of pentagon is 14.

2007-02-07 00:08:10 補充：
The 2nd question is tricky. You have to realise that he has is $33 $1 = $34 (purchase is 33 and the change is 1).So,let x be the no of $2 coins and y be the $5 conins2x+5y=34 ..........(1), x+y=11 .........(2)therefore, 7 $2 coins and 4 $5 coins

let the number of pentagons and the number of hexagons be X and Y
6Y + 5X = 100 ------------------- (1)
2(6Y) + 5X - 6(5) = 100
Y = (130 - 5X)/12 ----------------(2)
Sub (2) into (1),
(130 - 5X)/2 +5X = 100
5X = 70
X =14
Ann originally make 14 pentagons

2007-02-06 23:13:06 補充：
樓上錯o架 = =