Question

二元一次方程组要过程 x-5y/3+(x+y)/4=1 3(x-5y)+2(x+y)=4

Answer 1

3(x-5y)+2(x+y)=4
3X-15Y+2X+2Y=4
3X-13Y=4
X=(13Y+4)/3 代入 x-5y/3+(x+y)/4=1
得：(13Y+4)/3-5Y/3+[(13Y+4)/3+Y]/4=1
16Y+20/3=4
Y=-1/6
X=(13Y+4)/3=-37/18

Answer 2

x-5y/3+（x+y）/4=13（x-5y）+2（x+y）=4
x-5y/3+x/4+y/4=13x-65y+2x+2y=4 -17y+63y=4 y=2/23 15x-34/23=48
15x=48+34/23
15x=1104/23+34/23
15x=1138/23
x=1138/345
x-5y/3+x/4+y/4=4
13x-65y+2x+2y=4
5x/4+y/4-5y/3=4
5x/4+3y/12-20y/12=4
5x/4-17y/12=4
15x/12-17y/12=4
15x-17y=48
13x-65y+2x+2y=4
15x-63y=4