中二簡單simultaneous equation~~~

1.2x=7y-12......(1)
2y+3x=7......(2)

(1), x=(7y-12)/2......(3)

Sub. (3) into (2),
2y+3[(7y-12)/2]=7
4y+3(7y-12)=14
4y+21y-36=14
25y=50
y=2

Sub. y=2 into (3),
x=(14-12)/2=1

Therefore, the solution is x=1 and y=2.

2.3x-2y=7......(1)
4x-3y=-2........(2)

(1), x=(7+2y)/3......(3)

Sub. (3) into (2),
4[(7+2y)/3]-3y=-2
4(7+2y)-9y=-6
28+8y-9y=-6
-y=-34
y=34

Sub. y=34 into (3),
x=(7+68)/3=25

Therefore, the solution is x=25 and y=34.

3.-x+y+17=x+2y=16

-x+y+17=16......(1)
x+2y=16.....(2)

(2), x=16-2y.....(3)

Sub. (3) into (1),
-(16-2y)+y+17=16
-16+2y+y+17=16
3y=15
y=5

Sub. y=5 into (3),
x=16-2(5)=6

Therefore, the solution is x=6 and y=5.

1.2x=7y-12
x= (7y-12)/2
2y+3[(7y-12)/2]=7
4y+21y-36=14
25y=50
y=2

x=[7(2)-12]/2
x=1

2. 3x-2y=7
x=(7+2y)/3
4[(7+2y)/3]-3y= -2
4(7+2y)-9y=-6
28+8y-9y=-6
-1y=-34
y=34

x=[7+2(34)]/3
x=25

3. -x+y+17=16
x+2y=16

-x+y+17=16
y=16+x-17
x+2(16+x-17)=16
x+32+2x-34=16
3x=18
x=6
y=16+(6)-17
y=5

1.
2x=7y-12 .....(1)
2y+3x=7 .....(2)
From (1), x=(7y-12) / 2 ...(3)
Substitute (3) into (2),
2y + 3[(7y-12) / 2] = 7
4y + 3(7y-12) = 14
4y + 21y - 36 = 14
25y = 50
y=2
Subsitute y=2 into (1),
2x=7(2)-12
x=1
The solution is x=1 and y=2
2.
3x-2y=7....(1)
4x-3y=-2....(2)
From (1), x=(7+2y)/2 ....(3)
Substitute (3) into (2),
4[(7+2y)/2] - 3y = -2
8(7+2y) - 6y = -4
56 + 16y -6y = -4
10y= -60
y= -6
Substitute y=-6 into (1),
3x-2(-6)=7
3x+12=7
x=-5/3
The solution is x=-5/3 and y=-6
3.
-x + y + 17 = 16 ... (1)
x + 2y = 16 ...(2)
From (2), x=16-2y...(3)
Substitute (3) into (1),
-(16-2y) + y + 17 = 16
-16+2y + y + 17 = 16
3y=15
y=5
Substitute y=5 into (3),
x= 16 - 2(5)
= 16-10
= 6
The solution is x=6 and y=5

2009-10-07 21:27:19 補充：
correction:
From (1), x=(7+2y)/2 ....(3) < should be x=(7+2y)/3 ...(3)
Substitute (3) into (2),
4[(7+2y)/3] - 3y = -2
4(7+2y) - 9y = -6
28+8y-9y = -6
-y = -34
y=34

Substitute y=34 into (1),
3x-2(34)=7
3x=75
x=25
The solution is x=25 and y=34

2009-10-07 21:29:29 補充：
correction:
From (1), x=(7+2y)/2 ....(3) < should be x=(7+2y)/3 ...(3)
Substitute (3) into (2),
4[(7+2y)/3] - 3y = -2
4(7+2y) - 9y = -6
28+8y-9y = -6
-y = -34
y=34

Substitute y=34 into (1),
3x-2(34)=7
3x=75
x=25
The solution is x=25 and y=34