It is fast to solve the system of linear equations of three variables X + y + Z = 12, x + 2Y ― z = 6, 3x ― y + Z = 10

It is fast to solve the system of linear equations of three variables X + y + Z = 12, x + 2Y ― z = 6, 3x ― y + Z = 10

z=12-x-y
Then: x + 2y-z = 2x + 3y-12 = 6 (1)
3x-y+z=2x-2y+12=10 (2)
(1) 2
5Y = 20 is y = 4
Substituting y = 4 into (1) yields x = 3
So z = 12-x-y = 5

To solve the equations: 1) 2x + y = 2 2) 3x-2y = 10, x =, y=

1）2x+y=2
2）3x-2y=10
1) Multiply 2 to get 4x + 2Y = 4, 3)
2) 7x = 14
x=2
Substituting 1)
y=2-2*2=-2
x=2 y=-2

Solve the equations: x + y2 = 3x − 2Y = 10 + 6x + Y4

According to the meaning of the question, we get x + y2 = 3x − 2yx + y2 = 10 + 6x + Y4, sort out x − y = 0 (1) 4x − y = − 10 (2), from (1) - (2), we get x = - 103 (3). Substituting (3) into (1), we get y = - 103, so the solution of the original equations is x = − 103y = − 103