解三元一次方程组：x/2=y/3=z/5,2x+y+3z=88急用,
问题描述：
解三元一次方程组：x/2=y/3=z/5,2x+y+3z=88
急用,
答
令x/2=y/3=z/5=k
则x=2k,y=3k,z=5k
带入2x+y+3z=88
4k+3k+15k=88
22k=88
k=4
所以x=2k=8
y=3k=12
z=5k=20