When solving the equation x + y + 2Z = 7 2x + 3y-z = 12 3x + 2Y + Z = 13, the unknowns should be eliminated first () we get the system of bivariate first-order equations with respect to ()

5x+5y=25 and 5x+7y=31
Results: y = 3, x = 2
Bring x = 2, y = 3 into the original equation (any one)
Results: z = 1
So: x = 2, y = 3, z = 1

Solving equations X: y = 5:3, Y: z = 7:3, 2x-y-z = 34 (process)

x: If y = 5 / 3, then x = 5 / 3 y. (1)
y: Z = 7 / 3 then z = 3 / 7 y. (2)
Substituting the last formula, 10 / 3 y - Y - 3 / 7, y = 34
It is concluded that 40 / 21 y = 34 then y = (34 * 21) / 40
Substituting y value into formula (1) (2), X and Z can be obtained

Solving the system of three variable first order equations {X / 7 = Y / 10 = Z / 5,2x + 3y-2z = 34

The substitution method is the simplest
Let X / 7 = Y / 10 = Z / 5 = t, then x = 7T, y = 10t, z = 5T
Substituting 2x + 3y-2z = 34
2×7t+3×10t-2×5t=34
34t=34
T=1
x=7t=7 y=10t=10 z=5t=5
x=7 y=10 z=5

2X + y = 4, x + 3Z = 1, x + y + Z = 7 to solve the ternary system of first order equations

2x+y=4（1）,x+3z=1（2）,x+y+z=7（3）
From (2), x = 1-3z (4)
Replace (4) with (1)
2（1-3z）+y=4
y=2+6z（5）
Replace (4) (5) with (3)
（1-3z）+（2+6z)+z=7
The solution is Z = 1
Substituting z = 1 into (4), (5) yields
x=-2,y=8
Note: adhere to the principle of elimination, three into two, two into one, and finally solve the first order equation of one variable

Solving the system of three variable first order equations: X / 2 = Y / 3 = Z / 5, 2x + y + 3Z = 88 Urgent,

Let X / 2 = Y / 3 = Z / 5 = K
Then x = 2K, y = 3k, z = 5K
Bring in 2x + y + 3Z = 88
4k+3k+15k=88
22k=88
K=4
So x = 2K = 8
y=3k=12
z=5k=20

x:y:z=7:10:5 ① 2x+3y=44 ②

x: Y = 7:10, x = (7 / 10) y
In 2x + 3Y = 44, y = 10, so x = 7, z = 5

(2x-y)^7(y-2x)^5(y-2x)^4

(2x-y)^7(y-2x)^5(y-2x)^4
=(2x-y)^7[-(2x-y)^5](2x-y)^4
=-(2x-y)^(7+5+4)
=-(2x-y)^16

Solving the equations 2x + Y-Z = 7 x + y + Z = 1 2x-y-z = 5 Elimination

Formula 1 - 3 gives y = 1
Formula 2 + 3 gives x = 2
We know that if x and y are brought into any equation, we can get z = - 2

Solving the system of three variable linear equations {2x + Y-Z = 2 {x + 2Y = 5 {X-Y + 2Z = - 7

2x+y-z=2 （1）
x-y+2z=-7 （2）
x+2y=5 （3）
2 (1) + (2)
5x+y=-3 （4）
2 (4) - (3)
9x=-11
x=-11/9
5 (3) - (4)
9y=28
y=28/9
By substituting (1), we can get the following results:
z=4/3
Qi
x=-11/9
y=28/9
z=4/3

To solve the system of three variable linear equations: X-Y + Z = 7 2x-y-z = 0 x + y = - 1

X-Y + Z = 7 (1) 2x-y-z = 0 (2) (1) + (2): 3x-2y = 7 (3) x + y = - 1y = - 1-x substituting y = - 1-x into 3x-2y = 7 gives 3x-2 (- 1-x) = 73x + 2 + 2x = 75X = 5x = 1y = - 1-x because 2x-y-z = 0z = 2x-y = 2 * 1 - (- 2) = 4x = 1, y = - 2, z = 4

When solving the equation x + y + 2Z = 7 2x + 3y-z = 12 3x + 2Y + Z = 13, the unknowns should be eliminated first () we get the system of bivariate first-order equations with respect to ()