x^2-3y=9 x+y=3

To solve the system of equations:

1. \( x^2 - 3y = 9 \) (Equation 1)
2. \( x + y = 3 \) (Equation 2)

we can start by solving Equation 2 for one of the variables. Let's solve for \( y \):

\[ y = 3 - x \]

Now we can substitute this expression for \( y \) into Equation 1:

\[ x^2 - 3(3 - x) = 9 \]

Simplifying this gives:

\[ x^2 - 9 + 3x = 9 \]

Combining like terms, we have:

\[ x^2 + 3x - 18 = 0 \]

Now we can factor this quadratic equation. We need two numbers that multiply to -18 and add to 3. These numbers are 6 and -3.

So we can factor it as:

\[ (x + 6)(x - 3) = 0 \]

Setting each factor equal to zero gives:

\[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \] \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

Now we can find the corresponding \( y \) values for each \( x \).

1. If \( x = -6 \): \[ y = 3 - (-6) = 3 + 6 = 9 \] So, one solution is \( (-6, 9) \).

2. If \( x = 3 \): \[ y = 3 - 3 = 0 \] So, another solution is \( (3, 0) \).

Thus, the solutions to the system of equations are:

\[ (-6, 9) \quad \text{and} \quad (3, 0) \]