Question

x+y/2=z+x/3+y=z/4,...x+y+z=27.解方程组.

Answer 1

设(x+y)/2=(x+z)/3=(y+z)/4=k

则x+y=2k x+z=3k y+z=4k

代入x+y+z=27

2k+3k+4k=27 k=3

所以 x+y= 6 (1)

x+z=9 (2)

y+z=12 (3)

(1)+(2)+(3) 2(x+y+z)=27 x+y+z=27/2 (4)

(4)-(1) z=15/2

(4)-(2) y=9/2

(4)-(3) x=3/2

即为所求