F.3 Maths: elimination

Perrie

2008-02-28 6:55 am

solve the following simultaneous linear equations by the method of elimination
1. {5(3r - 1) + 4(3 + 2s) = 6
{5(4 - s) - 6(2r + 1) = 12
2. {x/3 - y/5 = 7
{x/4 + y/3 = - 2
3. 2x - 5 / 3 = x+3y /-2 = 1-y /3
*因為不能打一個大的 { , 所以用了兩個小的

回答 (1)

¦ý´²

2008-02-28 7:23 am

✔ 最佳答案

1)
5(3r - 1) + 4(3 + 2s) = 6 => 15r+8s=-1---(1)
5(4 - s) - 6(2r + 1) = 12 => -12r-5s=-2 ---(2)
5*(1)+8*(2):
75r+40s-96r-40s=-21
-21r=-21
r=1 ---(3)
Sub. (3)into(1):
15(1)+8s= -1
s= -2

2)
x/3 - y/5 = 7 => 5x-3y=105 ---(1)
x/4 + y/3 = - 2 => 3x+4y= -24 ---(2)
4*(1)+3*(2):
20x-12y+9x+12y=348
29x=348
x=12 ---(3)
Sub. (3)into(1):
5(12)-3y=105
3y= -45
y= -15

3)
(2x - 5) / 3 = (x+3y) /-2 => -4x+10 = 3x+9y => 7x+9y=10 ---(1)
(2x - 5) / 3 = (1-y) /3 => 2x+y=6 ---(2)
(1)-9*(2):
7x+9y-18x-9y = -44
-11x=-44
x=4 ---(3)
Sub. (3)into(1):
7(4)+9y=10
9y=-18
y= -2

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