3x-2y+z=3 2x+y-z=4 4x+3y+2z=-10

3x-2y+z=3 2x+y-z=4 4x+3y+2z=-10

3x-2y+z=3（1） 2x+y-z=4（2）4x+3y+2z=-10（3）,
From (2), z = 2x + y-4 (4),
By substituting (4) into formula (1) (3), 3x-2y + (2x + y-4) = 3 is obtained, and 5x-y = 7 (5) is obtained,
4X + 3Y + 2 (2x + y-4) = - 10 is simplified to 8x + 5Y = - 2 (6),
From (5) (6), x = 1, y = - 2
Substituting (4) gives z = - 4
Therefore, the solution of the original ternary system of first order equations is x = 1, y = - 2, z = - 4

Merge category 6x? - 2x-4x? 2x + 5x + 1

6x²-2x-4x²+5x+1
=6x²-2x-4x²+5x+1
=6x²-4x²-2x+5x+1
=(6-4)x²+(-2+5)x+1
=2x²+3x+1

When solving the equation, it is not necessary to merge the same kind of terms () a 2x = 3x B 2x + 1 = 0 C 6x-1 = 5x D 4x = 2 + 3x

Choose B

Given that the solutions of the equations 4x + 2m = 3x + 1 and 5x + 2m = 6x-3 on X are the same, find the value of (2m + 2) ^ 2007 Given that the solutions of the equations 4x + 2m = 3x + 1 and 5x + 2m = 6x-3 on X are the same, find the value of (2m + 2) ^ 2007

The solutions of 4x + 2m = 3x + 1 and 5x + 2m = 6x-3 are the same
From 4x + 2m = 3x + 1, x = 1-2m
5x + 2m = 6x-3 G is reduced to x = 3 + 2m
So 3 + 2m = 1-2m
M = - 1 / 2
So (2m + 2) ^ 2007
=（-1/2×2+2）^2007
=1^2007
=1

(1) X + 7 = 4 (2) 5x = 4x + 3 (3) - 2x = 6 (4) 0.5x + 1 = 3 junior high school mathematics, to test the process, it is important to test,

(1) X + 7 = 4 x = 4-7x = - 3 left = - 3 + 7 = 4 right = 4 left = right, so x = - 3 (2) 5x = 4x + 3 x = 3 left = 3 * 5 = 15 right = 4 * 3 + 3 = 15 left = right, so x = 3 (3) - 2x = 6 x = - 3 left = - 2 * (- 3) = 6 right = 6, left = right, so x = - 3 (4) 0.5x + 1 = 3x = 2 / 0.5x = 4left = 0.5 * 4 +

Solving equations 2x+3y−4z＝−7 x−4y 3＝2y+3z 2＝2

The original equations are rewritten as follows:
2x+3y−4z＝−7①
x−4y
3＝2②
2y+3z
2＝2③ ，
X = 6 + 4Y is obtained from equation ②, which can be simplified by substituting into ①
11y-4z=-19④，
From ③, 2Y + 3Z = 4 5,
From 4 × 3 + 5 × 4, we can get: 33y + 8y = - 57 + 16,
∴y=-1．
Substituting y = - 1 into ⑤, z = 2. Substituting y = - 1 into ②, x = 2
Qi
x＝2
y＝−1
Z = 2 is the solution of the original equations

Solving equations 2x+3y−4z＝−7 x−4y 3＝2y+3z 2＝2

The original equations are rewritten as follows:
2x+3y−4z＝−7①
x−4y
3＝2②
2y+3z
2＝2③ ，
X = 6 + 4Y is obtained from equation ②, which can be simplified by substituting into ①
11y-4z=-19④，
From ③, 2Y + 3Z = 4 5,
From 4 × 3 + 5 × 4, we can get: 33y + 8y = - 57 + 16,
∴y=-1．
Substituting y = - 1 into ⑤, z = 2. Substituting y = - 1 into ②, x = 2
Qi
x＝2
y＝−1
Z = 2 is the solution of the original equations

By the system of equations x−2y+3z＝0 2x − 3Y + 4Z = 0, X: Y: Z is equal to 0______ .

x−2y+3z＝0 ①
2x−3y+4z＝0 ② ，
① X 2 - 2, y = 2Z,
① X = Z,
∴x：y：z=z：2z：z=1：2：1．
So the answer is: 1:2:1

By the system of equations x−2y+3z＝0 2x − 3Y + 4Z = 0, X: Y: Z is equal to 0______ .

x−2y+3z＝0 ①
2x−3y+4z＝0 ② ，
① X 2 - 2, y = 2Z,
① X = Z,
∴x：y：z=z：2z：z=1：2：1．
So the answer is: 1:2:1

By the system of equations x−2y+3z＝0 2x − 3Y + 4Z = 0, X: Y: Z is equal to 0______ .

x−2y+3z＝0 ①
2x−3y+4z＝0 ② ，
① X 2 - 2, y = 2Z,
① X = Z,
∴x：y：z=z：2z：z=1：2：1．
So the answer is: 1:2:1