Question

Решить систему уравнений: x+2y=3. x^2+3xy-y^2=3 подробно

Answer 1

Всё в документе............................

Answer 2

To solve the system of equations:

1. x + 2y = 3 ............(Equation 1)
2. x^2 + 3xy - y^2 = 3 ...(Equation 2)

We can use the method of substitution or elimination to find the values of x and y.

Method of Substitution: Step 1: Solve Equation 1 for x: x = 3 - 2y

Step 2: Substitute the value of x from Step 1 into Equation 2: (3 - 2y)^2 + 3(3 - 2y)y - y^2 = 3

Step 3: Simplify and solve for y: Expand the square: (3 - 2y)^2 = (3 - 2y)(3 - 2y) = 9 - 6y - 6y + 4y^2 = 9 - 12y + 4y^2

Substitute in Equation 2: 9 - 12y + 4y^2 + 3(3 - 2y)y - y^2 = 3

Now, distribute the 3: 9 - 12y + 4y^2 + 9y - 6y^2 - y^2 = 3

Combine like terms: 4y^2 - 6y^2 - y^2 - 12y + 9y + 9 = 3

Simplify: -3y^2 - 3y + 9 = 3

Step 4: Move all terms to one side of the equation: -3y^2 - 3y + 6 = 0

Step 5: Divide the entire equation by -3 to simplify: y^2 + y - 2 = 0

Step 6: Factor the quadratic equation: (y + 2)(y - 1) = 0

Step 7: Set each factor to zero and solve for y:

1. y + 2 = 0 => y = -2
2. y - 1 = 0 => y = 1

Now we have two potential solutions for y: y = -2 and y = 1.

Step 8: Substitute the values of y back into Equation 1 to find the corresponding values of x:

For y = -2: x + 2(-2) = 3 x - 4 = 3 x = 3 + 4 x = 7

For y = 1: x + 2(1) = 3 x + 2 = 3 x = 3 - 2 x = 1

So, the solutions to the system of equations are: x = 7, y = -2 x = 1, y = 1