Quick,Please!!!!!!!!!!

{ 3x – 3y = 1 ─── (1)
{ 2x + y = 4 ─── (2)

From (2), we obtain

y = 4 – 2x

Sub y = 4 – 2x into (1),

3x – 3(4 – 2x) = 1

3x – 12 + 6x = 1

9x = 13

x = 13/9

Sub x = 13/9 into y = 4 – 2x

y = 4 – 2(13/9) = 10/9

The method you used should be called "substitution" (代入法)

3x - 3y = 1 ... (1)
2x + y = 4 ... (2)

From (2),
y = 4 - 2x ... (3)

Sub (3) into (1)
3x - 3(4-2x) = 1
3x - 12 + 6x = 1
9x = 13
x = 13/9 [or 1 4/9 (1又4/9)]

Sub x = 13/9 into (3)
y = 4 - 2(13/9)
y = 10/9 [or 1 1/9 (1又1/9)]


For your information, there is another method called elimination (消元法)
3x - 3y = 1 ... (1)
2x + y = 4 ... (2)

(2) x 3
6x + 3y = 12 ... (3)

(1) + (3)
(3x-3y) + (6x+3y) = 1+12
9x = 13
x = 13/9

Sub x=13/9 into (1)
y = 4 - 2(13/9)
y = 10/9