hi wat method would i use and how would i solve this system of equations

x^2+xy=3
3x^2+2xy-2y^2=7

In your first equation, solve for y... you should have xy= -x^2+3... divide this by x... so y = x^2+3/x

plug what you solved for y into your second equation and then solve for x

To solve this system of equations, we will follow these steps:

1. Begin with the first equation: x^2 + xy = 3.

2. Solve for y in the first equation. To do this, you can rearrange the equation by subtracting x^2 from both sides:

xy = -x^2 + 3.

Now, divide both sides of the equation by x:

y = (-x^2 + 3) / x.

So, you have obtained an expression for y in terms of x.

3. Substitute this expression for y into the second equation: 3x^2 + 2xy - 2y^2 = 7.

Replace y with (-x^2 + 3) / x in the equation:

3x^2 + 2x((-x^2 + 3) / x) - 2((-x^2 + 3) / x)^2 = 7.

4. Simplify the equation. Distribute the 2x and simplify the squared term:

3x^2 - 2x^3 / x + 6x^2 / x^2 - 6 / x^2 = 7.

Combine like terms and simplify the fractions:

3x^2 - 2x^2 + 6 - 6 / x^2 - 2x^3 / x = 7.

Simplify further:

x^2 - 2x^2 + 6 - 6 / x^2 - 2x = 7.

5. Rearrange the equation to form a quadratic equation:

-x^2 - 2x^2 - 2x + x^2 - 6 = 7.

Consolidate like terms:

-2x^2 - 2x - 6 = 7.

Simplify:

-2x^2 - 2x - 13 = 0.

6. Solve the quadratic equation by factoring, completing the square, or using the quadratic formula. For this equation, you will need to use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a).

In our equation, a = -2, b = -2, and c = -13.

Substituting the values into the quadratic formula:

x = (-(-2) ± √((-2)^2 - 4(-2)(-13))) / (2(-2)).

Simplify and compute to find the values of x:

x = (2 ± √(4 - 104)) / (-4).

x = (2 ± √(-100)) / (-4).

7. Since the square root of a negative number is not a real number, this equation has no real solutions. Therefore, there are no real solutions to the given system of equations.

Thus, the system of equations is unsolvable in the real number domain.