If the solutions X and y of the equations 3x-2y = K + 2, x + 6y = K-6 are opposite to each other, then K= I'm about to fall asleep. I'm so tired

If the solutions X and y of the equations 3x-2y = K + 2, x + 6y = K-6 are opposite to each other, then K= I'm about to fall asleep. I'm so tired

x. If y is opposite to each other, then x = - Y
5x=k+2
-5x=k-6
The solution is k = 2

Equations x + 4 = 7, xy = 12

x +4=7
→x=7-4
→x=3
Substituting x = 3 into xy = 12 leads to
y=4
The solution of the equations is x = 3, y = 4

To solve the equations, XY / x + y = 4 / 5, YZ / y + Z = 20 / 9, XZ / Z + x = 5 / 6
X + y is the denominator, and so is y + Z + X

XY / x + y = 4 / 5, then x + Y / xy = 5 / 4, then 1 / y + 1 / x = 5 / 4. ① YZ / y + Z = 20 / 9, then y + Z / YZ = 9 / 20, then 1 / Z + 1 / Y = 9 / 20. ② XZ / Z + x = 5 / 6, then Z + X / XZ = 6 / 5, then 1 / x + 1 / z = 6 / 5. ③ ① - ②, we get 1 / X-1 / z = 5 / 4-9 / 20, then 1 / X-1 / z = 4 / 5. ③ + ④, we get 2 / x = 2, we get x = 1 substituting ③, we get 1 + 1 / z = 6

Solving the system of equations x + y = 3, xy = 9 / 4 + 2Z

This is a ternary system of equations. If you want to solve it, you must have three equations. So now there must be infinite solutions

The solution of the equations {X / 24 + Y / 18 = 55 / 65 {Y / 16 + X / 8 = 3 / 2}~

x/24+y/18=55/65（1）
y/16+x/8=3/2（2）
（1）-（2）×1/3
y/18-y/48=55/65-1/2
y[30/（18×48）]=45/130
y=648/65
Substituting (2)
x/8=3/2-81/130
x/8=114/130
x=456/65
x=456/65
y=648/65

X / 24 + Y / 18 = 55 / 66, X / 8 + Y / 16 = 3 / 2, solve the equations

x/24+y/18=55/66 (1)
x/8+y/16=3/2 (2)
(1)×3-(2)
x/8+y/6-x/8-y/16=5/2-3/2
(1/6-1/16)y=1
(5/48)y=1
therefore
y=48/5
x=(3/2-y/16)*8=36/5

Y / 3-x + 1 / 6 = 3, 2 (X-Y / 2) = 3 (x + Y / 18)

y/3=3-1/6+x 2x-y=3x+y/6
y=9-0.5+3x -y=x+y/6
y=8.5+3x x=-7/6y
y=8.5+3(-7/6y) x=-7/6*9/17
y=8.5-3.5y x=-63/102
4.5y=8.5 x=-21/34
y=45/85
y=9/17

All integer solutions of Diophantine equation 2 (x + y) = XY + 7 are______ ．

(2x + 2x + 2Y + 2x + 2x + 2x + 2Y + 2y-4 = 3x (2-y) + 2 (Y-2) (2-y) = 3 (X-2) (2-2) (2-y) = 3 (X-2) when X-2 = 1, 2-y = 3 solution: x = 3: x = 3, y = 3: x = 3, y = 3 solution: x = 3, y = 3: x = 3, y = 1 (2) when X-2 = 3 = 2 = 3, 2-y = 1, 2-y = 1, 2-y = 1 solution: x = 5, and y = 1 (2) when X-2 = -2 = - 1, 2-2-2-y = - 2-2-2, 2-2-y = - 2-2-y = - 3, 2-2-y = - 3, 2-y = - 3 solution: x = 3 solution: x = 1: x = 1: x = 1: x = 1, y = 5, 1) (1, 5) ）（-1，3）

1. It is known that the solution of the system of equations x + y = m, 5x + 3Y = 31 about X and Y is nonnegative, so we can find the solution of integer M
2. It is known that the solutions of the equations X-Y = 5A + 1, x + y = 3A + 9 are positive
(1) The value range of a
(2) Simplify 4A + 5-a-4

1) By subtracting three times of Formula 1 from formula 2, we get 2x = 31-3m, so x = (31-3m) / 2,
Substituting y = (5m-31) / 2,
Because X and y are non negative numbers, 31-3m > = 0, 5m-31 > = 0,
The solution is 31 / 50,
The solution is - 5 / 4

Y = 1 / 2 + radical (2x-1) + radical (1-2x), find the square of X - XY + the square of Y

Because 2x-1 ≥ 0,1-2x ≥ 0,
So 2x = 1,
So x = 1 / 2,
So y = 1 / 2,
So x ^ 2-xy + y ^ 2
=(1/2)^2-(1/2)(1/2)+(1/2)^2
=1/4.