If the equation system 2x + 3Y = k 3x + 2Y = K-2, if the equation system 2x + 3Y = k 3x + 2Y = K-2, the sum of the solutions X and y of the equation system 2x + 3Y = k 3x + 2Y = K-2 is 8, find the value of K!

If the equation system 2x + 3Y = k 3x + 2Y = K-2, if the equation system 2x + 3Y = k 3x + 2Y = K-2, the sum of the solutions X and y of the equation system 2x + 3Y = k 3x + 2Y = K-2 is 8, find the value of K!

The addition of two equations 5x + 5Y = 2k-2
5(x+y)=2K-2
k=21

Try to find the value of the algebraic formula (2x + y) ^ 2-2 (2x + y) (X-Y) + (X-Y) ^ 2, where x and y are the solutions of the system of binary linear equations 2x-3y = 13,3x + 2Y = 0

x. Y is the solution of the bivariate linear equations 2x-3y = 13,3x + 2Y = 0
x=2,y=-3
So (2x + y) ^ 2-2 (2x + y) (X-Y) + (X-Y) ^ 2
=【(2x+y)-(x-y)】^2
=(x+2y)^2
=(2-2×3)^2
=16

2x-3y = 7 2x + 3Y = 1

(1) + 2
4x=8
∴x=2
Substituting x = 2 into (1) yields
4-3y=7
∴y=-1
The solution of the system of equations is
X=2
y=-1

Given that the system of bivariate linear equations is 4x + 3Y = 15,2x-y = 17, then the power of (x + 2Y) 2006 =?

4x+3y=15
2x-y=17
Subtracting two formulas: 2x + 4Y = - 2
That is: x + 2Y = - 1
So (x + 2Y) ^ 2006 = 1

Equations 4x−y＝k If the values of X and Y in 2x + 3Y = 1 are equal, then K=______ .

Solving equations
4x−y＝k
2X + 3Y = 1
x=3k+1
14、y=2−k
7，
∵x=y，
∴3k+1
14=2−k
7，
The solution is k = 3
5．
So the answer is 3
5．

Given the system of equations: 4x + 3Y = 6-k, 2x = k, the solution x is equal to the value of Y, so find the value of K Given the system of equations: 4x + 3Y = 6-k, 2x = k, the solution x is equal to the value of Y, so find the value of K

From x = y
7X=6-K
K = 2x
Substituting into the above formula
7X=6-2X
That is, 5x = 6, x = 5 / 6
K=2X=5/3

Solve the system of three variable linear equations 2x = 3Y ① 4Y = 5Z ② x + y + Z = 66 ③ the second 3x-y + Z = 4 ① 2x + 3y-z = 12 ② x + y + Z = 6

By substituting (1 + 2 / 3 + 8 / 15) x = 66 (11 / 5) x = 66, x = 30, y = 20, z = 16, ① + ②, 5x + 2Y = 16, ④ + ③, 3x + 4Y = 18, ⑤, ④ - ⑤, X = 2, and x = 2, respectively

Simplification evaluation: x + 2 (3y2-2x) - 4 (2x-y2), where | X-2 | + (y + 1) 2 = 0

The original formula = x + 6y2-4x-8x + 4y2 = - 11x + 10y2,
∵|x-2|+（y+1）2=0，
∴x=2，y=-1，
Then the original formula = - 22 + 10 = - 12

The following equations are solved: (1) {x-3y = 2x + y = 18 (2) {2A + B = 0 4A + 3B = 6 (3) {x-3y + 20 = 0 3x + 7y-100 = 0 (4) {2y-8 = - X Next to the fourth question, 4x + 3Y = 7 must be solved within today,

(1) ① + 3 × ② gives 7x = 54
X=8
From ①, y = (X-2) / 3 = 2
(2) 1 × 3 - 2, 2A = - 6
a=－3
Then B = - 2A = 6
(3) 3 × ① - ② gives - 16y = - 160
y=10
x=3y-20=10
(4) 5 y = 25
Y=5
x=8-2y=－2

Solving equations 2x−7y＝8 3x−8y−10＝0 ．

The original equations are changed into:
2x−7y＝8      ①
3x−8y＝10    ② ，
① X 3 - 2 × 2
-5y=4，
y=-4
5，
By substituting ①, we can get the following results:
X=6
5，
Qi
x＝6
Five
y＝−4
5 ．