急!!!F.2 MATHS[代入消元法]

5x + 3y + 6 = 0-----(1)
3x - 2y - 23 = 0-----(2)
(1)×2,
2(5x + 3y + 6) = 0×2
10x + 6y + 12 = 0 -----(3)
(2)×3,
3(3x - 2y - 23) = 0×3
9x - 6y - 69 = 0-----(4)
(3)+(4),
(10x + 6y + 12) + (9x - 6y - 69) = 0+0
10x + 6y + 12 + 9x - 6y - 69 = 0
19x - 57 = 0
19x = 57
x = 3
Sub x=3 into (1)
5x + 3y + 6 = 0
5(3) + 3y + 6 = 0
15 + 3y + 6 = 0
3y = - 21
y = -7
∴x = 3 y = -7


3x + 2y = x - 6 = 0

3x + 2y = 0-----(1)
x - 6 = 0-----(2)

From(2), x - 6 = 0
x = 6
Sub x = 6 into (1)
3x + 2y = 0
3(6) + 2y = 0
18 + 2y = 0
2y = -18
y = -9
∴x = 6 y = -9


2(x+y) - 3x = 4-----(1)
3(x- y)+ 5y = 12-----(2)

(1), 2(x+y) - 3x = 4
2x + 2y - 3x = 4
2y - x = 4-----(3)
(2), 3(x- y)+ 5y = 12
3x - 3y + 5y = 12
3x + 2y = 12-----(4)
(4)-(3),
(3x + 2y) - (2y - x) = 12 - 4
3x + 2y - 2y +x = 8
4x = 8
x = 2
Sub x=2 into (3)
2y - x = 4
2y - 2 = 4
2y = 6
y = 3
∴x = 2 y = 3

其實我唔係好知咩係代入消元法,
所以唔知你係咪要依個,
但係個答案一定岩!

我唔係好清楚咩係代入消元法,
但我諗應該係聯立方程計返x同y

Solution:
1.
5x + 3y + 6 = 0.......(1)
3x - 2y - 23 = 0.........(2)

From(1),
x= (-3y- 6)/5...........(3)

Sub (3) into (2),
3[(-3y- 6)/5] - 2y - 23 = 0 {成條式 * 5}
15(-3y - 6) - 10y - 115 = 0
-55y - 205 = 0
y = - 41/11...........(4)

Sub (4) into (3),
x = [-3(- 41/11)- 6]/5
x = 57/55



2.
3x + 2y = x - 6 = 0
即
3x + 2y = x - 6..........(10)
x - 6 = 0.............(11)

From (11),
x = 6

Sub x= 6 into (10),
3(6) + 2y = 6 - 6
y = - 9


3.
2(x+y) - 3x = 4
3(x- y)+ 5y = 12
即
2x + 2y -3x = 4
3x - 3y + 5y = 12
即
2y - x = 4..........(20)
8x - 3y = 12......(21)

From(20),
x = 2y - 4..........(22)

Sub (22) into (21)
8(2y - 4) - 3y = 12
13y = 44
y = 44/13.........(23)

Sub(23) into (22)
x = 2(44/13) - 4
x = 36/13

我不肯定答案必對,但希望幫到你.