X+2y+4z=31
5x+y+2z=29
3x-y+z=10
|1 2 4|
|5 1 2|=1-20+12-12+2-10=-27
|3 -1 1|
Δ_(x)=
|31 2 4|
|29 1 2| =31-116+40-40+62-58=-81
|10 -1 1|
Δ_(y)=
|1 31 4|
|5 29 2| =29+200+186-348-20-155=-108
|3 10 1|
Δ_(z)=
|1 2 31|
|5 1 29| = 10-155+174-93+29-100=-135
|3 -1 10|
x=( Δ_(x))/ Δ=(-81)/(-27)=3,
y=( Δ_(y))/ Δ=(-108_/(-27)=4,
z=( Δ_(z))/ Δ=(-135_/(-27)=5.
Ответ: (3; 4; 5).