二元一次方程 28+x-3y=8x-4(y+5)/3(4(y+5)除以3) 5(x-y)- 2(1-x)=3y/2
问题描述:
二元一次方程 28+x-3y=8x-4(y+5)/3(4(y+5)除以3) 5(x-y)- 2(1-x)=3y/2
答
28+x-3y=8x-4(y+5)/3(4(y+5)除以3)
28+x-3y=8x-1
29-3y=7x
58-6y=14x
5(x-y)- 2(1-x)=3y/2
5x-5y- 2+2x=3y/2
7x-5y- 2=3y/2
14x-10y-6=3y
14x=13y+6
58-6y=13y+6
52=19y
y=52/19
x=395/133