已知x+4y+z=24,2x+7y+2z=41,求x+y+z的值.
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已知x+4y+z=24,2x+7y+2z=41,求x+y+z的值.
![已知x+4y+z=24,2x+7y+2z=41,求x+y+z的值.](/uploads/image/z/2079321-33-1.jpg?t=%E5%B7%B2%E7%9F%A5x%2B4y%2Bz%3D24%EF%BC%8C2x%2B7y%2B2z%3D41%EF%BC%8C%E6%B1%82x%2By%2Bz%E7%9A%84%E5%80%BC%EF%BC%8E)
由已知得
x+4y+z=24
2x+7y+2z=41
设x+y+z=m(x+4y+z)+n(2x+7y+2z)=(m+2n)x+(4m+7n)y+(m+2n)z
比较系数,得
m+2n=1
4m+7n=1,解得
m=−5
n=3
∴x+y+z=m(x+4y+z)+n(2x+7y+2z)=-5×24+3×41=3.
x+4y+z=24
2x+7y+2z=41
设x+y+z=m(x+4y+z)+n(2x+7y+2z)=(m+2n)x+(4m+7n)y+(m+2n)z
比较系数,得
m+2n=1
4m+7n=1,解得
m=−5
n=3
∴x+y+z=m(x+4y+z)+n(2x+7y+2z)=-5×24+3×41=3.
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