救命啊﹗﹗﹗數學吾識﹗大家救我

1) Let x and y be the two numbers
x+ y = 27
as per your instruction, x-2y =3 therefore x = 3+2y
to substitue: (3+2y) + y = 27
therefore: 3 + 3y = 27
y= 8//
x= 19//

2) Let x and y be the two numbers
x - y = 72
as per your instruction: x= 5 y
to substitute: 5y-y =72
4y= 72
y= 18//
therefore: x = 90//

3) Let x and y be the two numbers
x + y = 63
as per your instruction: y-x =7, y=7+x
to substitute: x+(7+x) =63
2x= 63-7
x=28//
therefore: y= 35//

The first question

The sum of two numbers is 27 and one of them exceeds twice the other by 3.

let X be the first number and Y be the second number

X+Y = 27 -->1
Y= 2X+3 -->2

Sub 2 into 1

X+2X+3 = 27 --> 3
3X=24
X=8

Sub X=8 into 2

Y= 16+3
Y=19

Checking

19+8 = 27

Second question
One number is 5 times another and their difference is 72.

let the first number be X and second number be Y

X = 5Y -->1
Y-X=72 -->2

Sub 1 into 2

-4Y= 72
Y = -18

Sub Y=-18 into 2

-18-X=72
-X=90
X= -90

Checking
-90-(-18)

= -90+18
= -72

third question
The sum of two number is 63 and one of them exceeds the other by7

let the first number be X and second number be Y

X+Y =63 -->1
X-Y=7 -->2

From 2: X=7+Y -->3

Sub 3 into 1

7+Y+Y =63

2Y= 56
Y= 28

Sub Y=28 into 3
X=7+28
X= 35

Checking
35+28 = 63

2008-07-22 22:53:05 補充：
next time give more point...3 question for 5 point...i consider kind doing for u ^^" hehe