已知2x+5y+4z=15,7x+y+3z=14,3x+4y+2z=11.求4x+y+2z的值.

题目

已知2x+5y+4z=15,7x+y+3z=14,3x+4y+2z=11.求4x+y+2z的值.
列三元一次方程解

答案

a(2x+5y+4z)+b(7x+y+3z)=15a+14b
(2a+7b)x+(5a+b)y+(4a+3b)=15a+14b
2a+7b=4 5a+b=1 4a+3b=2
得到 a=1/11 b=6/11
代入1)式有
4x+y+2z=15×(1/11)+14×(6/11) =99/11=9