form 2 maths 急急急!!!20點~

2012-10-10 6:23 am
use the method of substitution to slove the following simultaneous equations

1. { 3x+y=13

x-y=13

2. {5x+7y=18
x=6=6

3. { 2x=7y-12
2y+3x=7

4. {3x-2y=7
4x+3y=7

5. -x+y+17=x+27=16

6. {x+2y=12
x-5y=-2

7. {3x+y=-1
9x+2y=1

8. {x-3y=-2
5x-2y=16

9. {2x-3y=1
4x-5y=5

10. { 7x-2y=2
3x-4y=-7

11. {5x+6y=-2
2x-3y=10

12. {3x+4y+2=0
6x+5y+7=0

13. {5x-2y+19=0
3x+4y+1=0

14. {2x+5y-8=0
9x+10y-11=0

show steps plz
thank you so much

回答 (1)

2012-10-10 10:24 pm
✔ 最佳答案
1.
3x + y = 13 ... [1]
x - y = 13 ... [2]

[2] :
y = x - 13 ... [3]

Put [3] into [1] :
3x + (x - 13) = 13
x = 13/2

Put into [3] :
y = -13/2


2.
What is meant by "x = 6 = 6" ?


3.
2x = 7y - 12 ... [1]
2y + 3x = 7 ... [2]

[1] :
x = (7y - 12)/2 ... [3]

Put [3] into [1]
2y + 3(7y - 12)/2 = 7
y = 2

Put into [3] :
x = 1


4.
3x - 2y = 7 ... [1]
4x + 3y = 7 ... [2]

[1] :
y = (3x - 7)/2 ... [3]

Put [3] into [2] :
x = 35/17

Put into [3] :
y = -7/17


5.
-x + y + 17 = 16 ... [1]
x + 27 = 16 ... [2]

From [2] :
x = -11

Put into [1] :
y = -12


6.
x + 2y = 12 ... [1]
x - 5y = -2 ... [2]

[2] :
x = 5y - 2 ... [3]

Put [3] into [1] :
(5y - 2) + 2y = 12
y = 2

Put into [3] :
x = 8


7.
3x + y = -1 ... [1]
9x + 2y = 1 ... [2]

[1] :
y = -3x - 1 ... [3]

Put [3] into [2] :
9x + 2(-3x - 1) = 1
x = 1

Put into [3] :
y = -4


8.
x - 3y = -2 ... [1]
5x - 2y = 16 ... [2]

[1] :
x = 3y - 2 ... [3]

Put [3] into [2] :
5(3y - 2) - 2y = 16
y = 2

Put into [3] :
x = 4


9.
2x - 3y = 1 ... [1]
4x - 5y = 5 ... [2]

[1] :
x = (3y + 1)/2 ... [3]

Put [3] into [2] :
[4(3y + 1)/2] - 5y = 5
y = 3

Put into [3] :
x = 5


10.
7x - 2y = 2 ... [1]
3x - 4y = -7 ... [2]

[1] :
y = (7x - 2)/2 ... [3]

Put [3] into [2] :
3x - 4(7x - 2)/2 = -7
x = 1

Put x = 1 into [3] :
y = 5/2


11.
5x + 6y = -2 ... [1]
2x - 3y = 10 ... [2]

From [2] :
y = (2x - 10)/3 ... [3]

Put [3] into [1] :
5x + 6(2x - 10)/3 = -2
x = 2

Put into [3] :
y = -2


12.
3x + 4y + 2 = 0 ... [1]
6x + 5y + 7 = 0 ... [2]

[1] :
x = (-4y - 2)/3 ... [3]

Put [3] into [2] :
[6(-4y - 2)/3] + 5y + 7 = 0
y = 1

Put into [3] :
x = -2


13.
5x - 2y + 19 = 0 ... [1]
3x + 4y + 1 = 0 ... [2]

[1] :
y = (5x + 19)/2 ... [3]

Put [3] into [2] :
3x + [4(5x + 19)/2] + 1 = 0
x = -3

Put into [3] :
y = 2


14.
2x + 5y - 8 = 0 ... [1]
9x + 10y - 11 = 0 ... [2]

[1] :
y = (8 - 2x)/5 ... [3]

Put [3] into [2] :
9x + [10(8 - 2x)/5] - 11 = 0
x = -1

Put into [3] :
y = 2
參考: micatkie


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