~F.2 maths X2

1.
2(x+1)=5(y-1)+11...............(1)
3(x+5)-2y=-1......................(2)
From (1),
2x + 2 = 5y - 5 + 11
2x = 5y + 4
x = ( 5y+4 ) / 2.....................(3)
Substitute (3) into (2)
3[( 5y+4 ) / 2 ) +5 ] -2y = -1
( 15y + 12 )/2 + 15 - 2y = -1
15y + 12 + 30 - 4y = -2
11y = -44
y = -4 //
Substitute y = -4 into (3),
x = [ 5(-4) + 4 ] / 2
x = -16 / 2
x = -8 //
The solution is x = -8, y = -4

2.
(x/3) - (y/4) =5 ...............(1)
3x +2y +11=5......................(2)
From (1),
4x - 3y = 60
4x = 3y + 60
x = ( 3y + 60 )/ 4....................(3)
Substitute (3) into (2)
3( 3y + 60 )/ 4 + 2y + 11 = 5
[(9y + 180) / 4] + 2y + 11 = 5
9y + 180 + 8y + 44 = 20
17y = -204
y = -12 //
Substitute y = -12 into (3)
x = 24 / 4
x = 6 //
The solution is x = 6, y = -12

第一題
2(x+1)=5(y-1)+11 <---equation1
3(x+5)-2y=-1 <---equation2

先拆曬其中1條equation , 我覺得equation2 比較好拆,所以我揀equation2黎拆
equation2 變成 3x-2y+16=0

之後將所以x以外嘅野整曬去右面
所以 3x-2y+16=0 會變成 x=(2y-16)/3
x=(2y-16)/3 <---equation3

將equation3 sub 落去 equation1 度
equation1 變成 2[(2y-16)/3+1]=5(y-1)+11
咁就計到 y=-4
sub y=-4 落去 equation1 or equation2 度
就會計到x=-6

第二題
x/3 - y/4 =3x+2y+11=5
將條equation拆成2條
即係 x/3 - y/4 =3x+2y+11 , 3x+2y+11=5
再用上面嘅方法
就會計到架啦

2007-03-13 21:02:39 補充：
第二題拆錯咗...係x/3 - y/4 =5 , 3x+2y+11=5