方程组x/7=y/10=z/5,x+2y+3z=84的解是?

问题描述:

方程组x/7=y/10=z/5,x+2y+3z=84的解是?

x/7=y/10=z/5 得 x=7z/5 y=2z
代入得 7z/5+4z+3z=84
z=10 则 x=14 y=20

令x/7=y/10=z/5=k,则x=7*k,y=10*k,z=5*k
x+2y+3z=7*k+2*10*k+3*5*k=84,解得k=2,所以
x=14
y=20
z=10

x/7=y/10=z/5=k
x=7k, y=10k, z=5k代入x+2y+3z=84,
7k+20k+15k=84
42k=84
k=2
x=7k=14;
y=10k=20;
z=5k=10.