聯立方程的問題

a
3x-2y=-1 --- 1
4x+5y=14 --- 2
我們可以用消元法
1x4 : (3x-2y)x4=-1x4 = 12x-8y=＝-4－－－３
2x3 : (4x+5y)x3=14x3 = 12x+15y=42－－－４

3-4 : (12x-8y)-(12x+15y)=(-4)-(42)
23y = -46
y = -2－－－5
5代入3:
12x-8(-2)=＝-4
12x+16=-4
12x=12
x=1

b
2x-y=-7 ----1
3x+2=7 ----2

1x2: 4x-2y ＝ -14－－－３

3+2 : (4x--2y)+(3x+2y)=(-14)+(7)
7x = -7
x = -1 －－－４
4代入2:
3(-1)+2y = 7
-3+2y = 7
2y = 10
y = 5

a
3x-2y=-1 --- 1
4x+5y=14 --- 2
1x4，12x-8y= -4-------3
2x3，12x+15y=42-----4
4-3
12x+15y-12x+8y=42+4
23y=46
y=2-----------5
5代入1
3x-2(2)= -1
x=1
所以x=1，y=2

b
2x-y=-7 ----1
3x+2=7 ----2

b果題唔知你係咪打少左個y(3x+2[y]=7 ----2)
所以我俾左兩個答案你

b(1)
2x-y= -7 ----1
3x+2=7 ----2
由2，x=5/3--------3
3代入2，
2(5/3)-y= -7
10/3-y= -7
y=31/3---------4
所以x=5/3，y=31/3

b(2)
2x-y=-7 ----1
3x+2y=7 ----2
1x2，4x-2y= -14-------------3
2+3
3x+2y+4x-2y=7-14
7x=-7
x=-1-------------4
4代入1
2(-1)-y= -7
y=5
所以x=-1，y=5

a) 4x(1) 12x-8y=-4 -----(3)

3x(2) 12x+15y=42-----(4)

(4)-(3) 23y=46

y=2

put y=2 into (3)

12x-16=-4

12x=12

x=1

b) 2x(1) 4x-2y=-14----(3)

(2)+(3) 7x=-7

x=-1

put x=-1 into (1)

2(-1)-y=-7

-2+7=y

y=5

a) 3x-2y=-1
2y = 3x + 1
y = 3x/2 + 1/2

Sub into 2
4x + 5(3x/2 + 1/2) = 14
11.5x + 2.5 = 14
x = 1
y = 3/2 + 1/2 = 2

b) 2x-y=-7
y = 2x + 7
Sub into 2
3x + 2(2x + 7) = 7
7x + 14 = 7
x = -1
y = 5