詳細的加減消元法的做法

加減消元法: (如果我無記錯的話-_-)
通常我們會考慮2條2元1次方程
Consider 1. A(x) + B(y) = C ----(1)
2. D(x) + E(y) = F -----(2)
where A,B,C,D,E,F are real numbers and x,y are variable (就是所謂的元)

In order to find the values of x and y ,
we need to eliminate either x or y in either one of the equations.
Now, we try to elminate y(消元),
(1)*(E/B) - (2) :
A(E/B)x + B(E/B)y - Dx - Ey = C(E/B) - F
A(E/B)x - Dx = C(E/B) - F
((AE-DB)/B)x = ((CE-FB)/B)
so , x = ((CE-FB)/B) / ((AE-DB)/B)
= (CE-FB) / (AE-DB)
Sub x = (CE-FB) / (AE-DB) into (1) to find y :
A((CE-FB) / (AE-DB)) + B(y) = C
y = (C - A((CE-FB) / (AE-DB))) / B

Above explanation seems to be complicated ,
Take an example :
1. 2x + 3y = 2 ---(1)
2. 3x + 4y = 3 ---(2)
To eliminate y :
(1) * (4/3) - (2) :
2(4/3)x+3(4/3)y-3x-4y = 2(4/3) - 3
(8/3)x-3x = 8/3-3
-(1/3)x = -(1/3)
x = 1
Sub x = 1 into (1),
we have y = 0