Question

решите систему уравнений

Answer 1

Решение в приложении.

Answer 2

Solving the System of Equations

To solve the system of equations, we will use the information provided in the search results. Let's examine each source and extract the relevant information.

From Source we have the following system of equations: - Equation 1: 2x - y = 7 - Equation 2: x^2 - xy = 12

From Source we have the following system of equations: - Equation 1: 7x + 8y = 2 - Equation 2: 34x - 6y = -7

From Source we have the following system of equations: - Equation 1: x + y = 6 - Equation 2: x^2 + y^2 = 18

From Source we have the following system of equations: - Equation 1: x + 2y = 4 - Equation 2: 7x - 3y = 5

Now, let's solve each system of equations one by one.

System of Equations from SourceEquation 1: 2x - y = 7

Equation 2: x^2 - xy = 12

To solve this system, we can use substitution or elimination method. Let's use the substitution method.

From Equation 1, we can express y in terms of x: y = 2x - 7

Substituting this value of y into Equation 2, we get: x^2 - x(2x - 7) = 12 x^2 - 2x^2 + 7x = 12 -x^2 + 7x = 12 x^2 - 7x + 12 = 0

Now, we can solve this quadratic equation. From Source we can see that the solutions are: x = 4 - 223 or x = 4 + 223

Substituting these values of x back into Equation 1, we can find the corresponding values of y: For x = 4 - 223, y = -2(2 + 23) = -52 For x = 4 + 223, y = -2(2 - 23) = 42

Therefore, the solutions to the system of equations from Source are: (4 - 223, -52) and (4 + 223, 42)

System of Equations from SourceEquation 1: 7x + 8y = 2

Equation 2: 34x - 6y = -7

To solve this system, we can use the elimination method. Let's eliminate y by multiplying Equation 1 by 6 and Equation 2 by 8:

6(7x + 8y) = 6(2) => 42x + 48y = 12 8(34x - 6y) = 8(-7) => 272x - 48y = -56

Adding these two equations together, we get: 42x + 48y + 272x - 48y = 12 - 56 314x = -44 x = -44/314 x = -0.14

Substituting this value of x back into Equation 1, we can find the value of y: 7(-0.14) + 8y = 2 -0.98 + 8y = 2 8y = 2 + 0.98 8y = 2.98 y = 2.98/8 y = 0.3725

Therefore, the solution to the system of equations from Source is: (x, y) = (-0.14, 0.3725)

System of Equations from SourceEquation 1: x + y = 6

Equation 2: x^2 + y^2 = 18

To solve this system, we can use the substitution method. Let's express x in terms of y from Equation 1: x = 6 - y

Substituting this value of x into Equation 2, we get: (6 - y)^2 + y^2 = 18 36 - 12y + y^2 + y^2 = 18 2y^2 - 12y + 18 = 0 y^2 - 6y + 9 = 0 (y - 3)^2 = 0

From this equation, we can see that y = 3. Substituting this value of y back into Equation 1, we can find the value of x: x + 3 = 6 x = 6 - 3 x = 3

Therefore, the solution to the system of equations from Source is: (x, y) = (3, 3)

System of Equations from SourceEquation 1: x + 2y = 4

Equation 2: 7x - 3y = 5

To solve this system, we can use the elimination method. Let's eliminate x by multiplying Equation 1 by 7 and Equation 2 by 1:

7(x + 2y) = 7(4) => 7x + 14y = 28 1(7x - 3y) = 1(5) => 7x - 3y = 5

Subtracting Equation 2 from Equation 1, we get: 7x + 14y - (7x - 3y) = 28 - 5 17y = 23 y = 23/17

Substituting this value of y back into Equation 1, we can find the value of x: x + 2(23/17) = 4 x + 46/17 = 4 x = 4 - 46/17 x = (68 - 46)/17 x = 22/17

Therefore, the solution to the system of equations from Source is: (x, y) = (22/17, 23/17)

In summary, the solutions to the given systems of equations are as follows: - System from Source (4 - 223, -52) and (4 + 223, 42) - System from Source (-0.14, 0.3725) - System from Source (3, 3) - System from Source (22/17, 23/17)

Please let me know if there's anything else I can help you with!