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(4x0.8)+(13x1.8)

Tom paid $26.6 for 17 stamps, consisting of $0.8 and $1.8 stamps. Find the no. of each kind of1 stamp.

Let x be the numbers of $0.8 stamps and y be the numbers of$1.8stamps.

17-x=y - (1)
{
0.8x+1.8y=26.6 - (2)

Sub.(1)into(2),
0.8x+1.8(17-x)=26.6
0.8x+30.6-1.8x=26.6
30.6-26.6=1.8x-0.8x
x=4
Sub. x=4 into (1)
17-x=y
17-4=y
13=y
There are 4 $0.8 and 13 $1.8 stamps.

Let x and y be number of $0.8 stamps and $1.8 stamps respectively.
0.8x + 1.8y = 26.6 <------ 1
x + y = 17 equal to x=17-y <----- 2
Sub 1 into 2
0.8(17-y) + 1.8y = 26.6
13.6 - 0.8y + 1.8y = 26.6
y=26.6-13.6
y=13 <-----3
Sub 3 into 2
x+13 = 17
x=4

2007-02-04 19:45:46 補充：
樓上果位 串錯 respectively 唷 改改佢

使用二元一次聯立方程來解答此問題。

Let x and y be $0.8 and $1.8 stamps representively.

｛x+y=17 (A)
｛0.8x+1.8y=$26.6 (B)

(A)*4 4x+4y=$68 (C)
(B)*5 4x+9y=$133 (D)
－－－－－－－－－－－－
(C)-(D) -5y=$-65

Put y=13 to the forumla (A),
x+13=17
x=4

So $0.8 stamps are 4 and $1.8 stamps are 13.

2007-02-04 19:42:24 補充：
您也可以使用一元一次方程來解答問題。Let z is the number of $0.8 stamps and (17-z) are the number of $1.8 stamps.0.8z 1.8(17-z)=26.60.8z 30.6-1.8z=26.60.8z-1.8z=26.6-30.6-z=-4z=4So $0.8 stamps are 4.

Tom paid $26.6 for 17 stamps, consisting of $0.8 and $1.8 stamps. Find the no. of each kind of1 stamp.

Let a be the number of $0.8 stamps, ( 17-a ) be the number of $1.8 stamps

0.8a + 1.8( 17-a ) = 26.6
0.8a + 30.6 - 1.8a = 26.6
-a = -4
a = 4 //

The number of $0.8 stamps is 4, the number of $1.8 stamps is 13 //

2007-02-04 19:42:40 補充：
You can also solve this question by using simultaneous equations in two unknowns~Let a be the number of $0.8 stamps, b be the number of $1.8 stampsa + b = 17....................(1)0.8a + 1.8b = 26.6........(2)From (1), a = 17 - b.....................(3)

2007-02-04 19:42:53 補充：
0.8( 17-b ) 1.8b = 26.613.6 - 0.8b 1.8b = 26.6-b = -13b = 13 //Substitute b = 13 into (3)a = 17 - 13a = 4 //The number of $0.8 stamps is 4, the number of $1.8 stamps is 13

2007-02-04 19:49:10 補充：
不過本人建議你用一元一次方程, 因為 (1) 用二元一次方程前, 一定要識計一元一次方程, .....而且二元一次方程係比較高 form D 的數, 唔鼓勵你 "未學行, 先學走"(2) 二元一次方程需要寫多D steps, 如果考試的時候, 較為waste time.....例如:substitute (1) into (2) , the solution is x = ?, y = ?........etc.