解1元二次方程,please

1)
x-y=8------(1)
7x+4y=67------(2)
From (1), x=8+y------(3)
Sub (3) into (2),
7(8+y)+4y=67
56+7y+4y=67
11y=11
y=1
Sub y=1 into (1), x=9

2)
5q-2=2q------(1)
3p-2p+24=2q------(2)
From (1),3q=2,
q=2/3
Sub q=2/3 into (2),
3p-2p+24=2(2/3)
p=4/3-24
p=-68/3

3)
1-3x=7y------(1)
2-2x+2y=7y------(2)
From (1),y=(1-3x)/7------(3)
Sub (3) into (2),
2-2x+2[(1-3x)/7]=7[(1-3x)/7]
2-2x+2[(1-3x)/7]=1-3x <---[Then times 7 on both sides]
14-14x+2(1-3x)=7(1-3x)
14-14x+2-6x=7-21x
x=-9
Sub x=-9 into (1),
1-3(-9)=7y
1-12=7y
y=-11/7

2011-09-01 20:44:18 補充：
Also, these are 2元1次方程

1)
x-y = 8 ...i
7x + 4y = 67 ...ii
7*ii,
7x-7y=56 ...iii
ii-iii,
11x=11
x=11
x=11 put into i
11-y=8
-y=-3
y=3
so (x,y)=(11,3)

2)
5q-2 = 2q = 3p - 2p + 24
5q-2=2q=p+24
5q-2=p+24 ...i
2q=p+24 ...ii
i,
p=5q-26...iii
iii put into ii
2q=5q-26+24
-3q=-2
q=2/3
q=2/3 put into iii
p=5(2/3)-26
p=--68/3
so (p,q)=(-68/3,2/3)

3) 1- 3x = 7y = 2- 2x + 2y
1-3x=7y ...i
7y=2-2x+2y ...ii
ii,
5y=2-2x
5y-2=-2x
x=(2-5y)/2 ...iii
iii put into i
1-3[(2-5y)/2]=7y
1-(6-15y)/2=7y
2-6-15y=14y
-4=29y
y=29/4
y=29/4 put into iii
x=[2-5(29/4)]/2
x=-137/8
so (x,y)=(-137/8,29/4)

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1) x-y = 8 --------------(1)
7x + 4y = 67 --------------(2)
(1)*4 + (2), 11x = 99 => x = 9
9-y = 8 => y = 1

2) 5q-2 = 2q => q = 2/3
3p - 2p + 24 = 2*2/3 => p = 68/3

3.)
1- 3x = 7y --------------(1)
7y = 2- 2x + 2y => 5y = 2 - 2x -------------(2)
(1)*2 - (2)*3, 2 - 6 = -y => y = 4
1 - 3x = 28 => x = -9