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secret13
@secret13
July 2022
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решите систему уравнений
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belozerova1955
Из 1 уравнения выразить х = 2+z-y и подставить во 2 и 3 уравнения
2(2+z-y) - y+4z=1
-(2+z-y)+6y+z=5
4+2z-2y-у+4z=1
-2-z+y+6y+z=5
6z-3y= -3 ( : 3) ⇒ 2z-у= -1
7y=7 ⇒ y=1
2z-у =-1
у=1
2z=-1+1
z=0
x=2+0-1
x=1
Ответ: ч=1; у=1; z=0
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Thanks 1
belozerova1955
Извини, не ч=1, а х=1
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Answers & Comments
2(2+z-y) - y+4z=1
-(2+z-y)+6y+z=5
4+2z-2y-у+4z=1
-2-z+y+6y+z=5
6z-3y= -3 ( : 3) ⇒ 2z-у= -1
7y=7 ⇒ y=1
2z-у =-1
у=1
2z=-1+1
z=0
x=2+0-1
x=1
Ответ: ч=1; у=1; z=0