中二Maths~唔識ar!!help me!! .

(i)*2: x-2y=1/3...........(iii)
(iii)+(ii): 4x/3=4/3
x=1
sub x=1 into (iii)
1-2y=1/3
-2y=-2/3
y=1/3

so x=1,y=1/3

聯立二元一次方程----加減法－－（這題是不能用代入法嗎？要用加減來消去？）
1.
x/2 - y = 1/6 .......(i)
x/3+ 2y = 1 .........(ii)
將(i)乘2，
得出x-2y=1/3........(iii)
將(ii)+(iii),
x+x/3=1/3+1
4x/3=4/3
x=1
將x=1代入(i)
即(1)/2-y=1/6
y=1/2-1/6
y=1/3
所以,x=1,y=1/3//

2.
x +6y = 3 .......... (1ii) x 3
x = 3 - 6y ............. (iii)

3.
(3 - 6y)/2 - y = 1/6 ....... sub (iii) into (i)
3 - 6y - 2y = 1/3
3 - 1/3 = 8y
8/3 = 8y
y = 1/3 .... (iv) [Answer]

4.
x/2 - 1/3 = 1/6 ...... sub (iv) into (1i)
3x - 2 = 1 ................
x = 1 ............... [Answer]

1. X/2-y=1/6
3X-6Y=1
3x=1+6y
x=1/3+2y

2. x/3+2y=1
x/3=1-2y
x=3-6y

代入返去
1/3+2y=3-6y
1/3+8y=3
8y=3-1/3
8y=9/3-1/3
8y=8/3
8y/8=8/3x1/8
y=1/3

再代入返去

x/2-1/3=1/6
x/2=1/6+2/6
x/2=1/2
x=1
or
x/3+2(1/3)=1
x/3+2/3=1
x/3=1-2/3
x/3=1/3
x=1

so...y=1/3 x=1
maybe it can help you

(i)......x/2-y=1/6
-y=1/6-1/2
-y=1/6-3/6
-y=-2/6
-y=-1/3
y=1/3
第二題我就唔識la~~~