~F.2 MATHS

用戶87351

2007-08-12 2:02 am

Solve the systems of equations by the method of elimination.
1) { 6x- (y-4)/3 =10-----(1)
y+ (3x-7)/4=1 -------(2)

Solve:(can use substitution or elimination)
2) { 5x+6y=1 ------(1)
(4/3)x - 2y=8/3 -------(2)

3) { 5/m +6/n=1-------(1)
4/3m -2/n =8/3 --------(2)

hint:let 1/m=x, 1/n=y

*Please show the process, thanks.

回答 (1)

用戶4872

2007-08-12 3:30 am

✔ 最佳答案

1)
6x- (y-4)/3 = 10 ---- (1)
y+ (3x-7)/4 = 1 ---- (2)
from (1), 18x - (y-4) = 30
18x - y = 26 ---- (3)
from (2), y+ (3x-7)/4 = 1
4y + (3x-7) = 4
4y + 3x = 11 ---- (4)
(3) - 6*(4),
18x - y - 6*(4y + 3x) = 26 - 6*(11)
18x - y - 24y - 18x = 26 - 66
-25y = -40
y = 40/25 = 8/5
4(8/5) + 3x = 11
32/5 + 3x = 11
32 + 15x = 55
15x = 23
x = 23/15

2)
5x+6y = 1 ---- (1)
(4/3)x - 2y = 8/3 ---- (2)
(1) + 3*(2),
5x + 6y + (4x - 6y) = 1 + 8
9x = 9
x = 1
5(1)+6y = 1
6y = 1 - 5
y = -4/6 = -2/3

3)
5/m + 6/n = 1 ---- (1)
4/3m -2/n = 8/3 ---- (2)

let 1/m = x, 1/n = y
Then,
5x + 6y = 1 ---- (1)
(4/3)x - 2y = 8/3 ---- (2)
From question no. 2,
x = 1 = 1/m
m = 1
y = -2/3 = 1/n
n = -3/2

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