Given that x, y satisfy that 2x + y in 2 is equal to 5x + 2Y in quarter, find the value of the algebraic formula 2x-3y + 3x + 2Y + 3

Given that x, y satisfy that 2x + y in 2 is equal to 5x + 2Y in quarter, find the value of the algebraic formula 2x-3y + 3x + 2Y + 3

(2x+y)/2=(5x+2y)/4
2(2X+Y)=(5X+2Y)
4X+2Y=5X+2Y
X=0
So x = 0, so the algebraic result is 19y / 7 + 3 / 7

The problem solving process of the system of three variable linear equations X / 7 = Y / 10 = Z / 5,2x + 3Y = 44, We have to be quick. We have to wait so hard

Because X / 7 = Y / 10, 10x = 7Y. X = 7 / 10Y
Substituting x = 7 / 10Y into 2x + 3Y = 44 can get y = 10
Substituting y = 10 into X / 7 = Y / 10 = Z / 5 gives x = 7, z = 5

The third order equation system of three variables X = 10 parts y = 5 parts Z ① 2x + 3Y = 44 ② 4Y + 5Z = 65 ③ ...

Let X / 7 = Y / 10 = Z / 5 = K
x=7k
y=10k
z=5k
2x+3y=44
2*7k+3*10k=44
44k=44
K=1
x=7k=7*1=7
y=10k=10*1=10
z=5k=5*1=5
The results show that x = 7, y = 10, z = 5 are consistent with 4Y + 5Z = 65
So x = 7, y = 10, z = 5

Solve the equations 2x + 3Y + 2Z = 19 3x + 2Y + 2Z = 17 2x + 2Y + 3Z = 13, find the values of X, y, Z

This is a junior high school mathematics problem, the solution is as follows
Let 2x + 3Y + 2Z = 19 be a, 3x + 2Y + 2Z = 17 be B, 2x + 2Y + 3Z = 13 be C: A, B and B, respectively
A is reduced to Z = (19-2x-3y) / 2
Substituting z = (19-2x-3y) / 2 into B and C, we get the following results
3X+2Y+2〈（19-2X-3Y）/2〉=17
2X+2Y+3〈（19-2X-3Y）/2〉=13
Then it is simplified as follows
X-Y=-2
2X+5Y=31
By solving the bivariate equation of first degree, the following results are obtained
X=3,Y=5
Then substituting X and Y into a, we get
2*3+3*5+2Z=19
Z = - 1
So, x = 3, y = 5, z = - 1

How to solve the equation system {2x + 3y-z = 18 {3x + 2Y + Z = 32 {x + 2Y + Z = 24

(3x+2y-z)+(x+2y+z)=2x=32-24=8 x=4
(2x+3y-z)+(3x+2y+z)=5x+5y=50 x+y=10 x=4 y=6
x+2y+z=24 x=4 y=6 z=8
The solution of the equation can be checked in the original equation after the result is obtained, so as to avoid miscalculation

Known equations 2x+3y＝10 If 3x + 2Y = 15, then x + y=______ .

2x+3y＝10①
3x+2y＝15② ，
① 5 (x + y) = 25,
Then x + y = 5
So the answer is: 5

The system of equations is 1y + 2Z + 1y + 2Z = 1y + 2Z

solution
3x+2y+z=9 ①
x+y+2z=0 ②
2x+3y-z=11 ③
① X 2: 6x + 4Y + 2Z = 18
④ - 2: 5x + 3Y = 18, ⑤
① The result is: 5x + 5Y = 20
⑥ - 5: 2Y = 2
ν y = 1, substituting y = 1 into 6 gives: x = 3
∴z=-2
∴x=3,y=1,z=-2

Solve the following equations: （1） 3x-2y=6 2x+3y=17 ；           （2） 4x-y-5=0 X 2+y 3=2 ．

（1）
3x-2y=6①
2x+3y=17② ，
① X 3 + 2 × 2: 13X = 52, i.e. x = 4,
By substituting x = 4 into ①, y = 3,
Then the solution of the equations is
X=4
y=3 ；
(2) The equations are as follows: 1
4x-y=5①
3x+2y=12② ，
① X 2 + 2: 11x = 22, i.e. x = 2,
By substituting x = 2 into ①, y = 3,
Then the solution of the equations is
X=2
y=3 ．

To solve the equations 4x−3y＝5 2x−y＝2

4x−3y＝5①
2x−y＝2② ，
From ②, y = 3x-2,
Substituting ①, 4x-3 (3x-2) = 5,
X=1
2．
Substituting into (2), y = - 1
So the solution of the equations is
x＝1
Two
y＝−1 ．

Solving equations 2x+y＝5 4x+3y＝7 ．

2x+y＝5  ①
4x+3y＝7② ，
① × 3 - ② is: 2x = 8,
The solution is: x = 4,
Substituting x = 4 into ①, 8 + y = 5,
The solution is: y = - 3,
Then the solution of the original equations is
x＝4
y＝−3 ．