Question

解下面二元一次方程（要过程）{x+y/3 + x-y/2 =6 {3(x+y)-2(x-y)=28

用加减消元法或者代入消元法（必须的）

Answer 1

(x+y)/3+(x-y)/2=6
2x+2y+3x-3y=36
5x-y=36
3x+3y-2x+2y=28
x+5y=28
26y=28*5-36
y=104/26
y=4
x=28-5y
x=8
所以，x=8，y=4