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vanmarkevich
@vanmarkevich
July 2022
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- x+y+z+u=5
I y+z+u+v=1
< z+u+v+x=2
I u+v+x+y=0
- v+x+y+z=4
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NNNLLL54
Verified answer
X+y+z+u =5
y+z+u+v=1
x + z+u+v=2
x+y + u+v=0
x+y+z + v=4
Сложим все уравнения, получим
4x+4y+4z+4u+4v=12
4(x+y+z+u+v)=12
x+y+z+u+v=3
(x+y+z+u)+v=3 ---> 5+v=3 , v= -2
x+(y+z+u+v)=3 ---> x+1=3 , x=2
y+(x+z+u+v)=3 ---> y+2=3 , y=1
z+(x+y+u+v)=3 ---> z+0=3 , z=3
u+(x+y+z+v)=3 ---> u+4=3 , u=-1
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Answers & Comments
Verified answer
X+y+z+u =5y+z+u+v=1
x + z+u+v=2
x+y + u+v=0
x+y+z + v=4
Сложим все уравнения, получим
4x+4y+4z+4u+4v=12
4(x+y+z+u+v)=12
x+y+z+u+v=3
(x+y+z+u)+v=3 ---> 5+v=3 , v= -2
x+(y+z+u+v)=3 ---> x+1=3 , x=2
y+(x+z+u+v)=3 ---> y+2=3 , y=1
z+(x+y+u+v)=3 ---> z+0=3 , z=3
u+(x+y+z+v)=3 ---> u+4=3 , u=-1