The system of equations given in the image is:

$y = 2x + 3 \quad \text{(Equation of a line)}$ $y = x^2 + 3x - 3 \quad \text{(Equation of a parabola)}$

To solve this system of equations, we can substitute $y = 2x + 3$ from the first equation into the second equation. This will allow us to find the points of intersection between the line and the parabola.

Step 1: Substitution

Substitute $2x + 3$ for $y$ in the quadratic equation:

$2x + 3 = x^2 + 3x - 3$

Step 2: Simplify

Move all terms to one side to set the equation to 0:

$x^2 + 3x - 3 - 2x - 3 = 0$

Simplify the terms:

$x^2 + (3x - 2x) - 3 - 3 = 0$ $x^2 + x - 6 = 0$

Step 3: Solve the quadratic equation

We now have the quadratic equation:

$x^2 + x - 6 = 0$

To solve this, we can factor it:

$(x - 2)(x + 3) = 0$

Thus, the solutions for $x$ are:

$x = 2 \quad \text{or} \quad x = -3$

Step 4: Find the corresponding $y$-values

Now substitute these $x$-values back into the linear equation $y = 2x + 3$ to find the corresponding $y$-values.

1. For $x = 2$: $y = 2(2) + 3 = 4 + 3 = 7$ So, one solution is $(2, 7)$.

2. For $x = -3$: $y = 2(-3) + 3 = -6 + 3 = -3$ So, the other solution is $(-3, -3)$.

Final Answer:

The solutions to the system are $(-3, -3)$ and $(2, 7)$, which corresponds to option C.

Would you like more details or have any questions? Here are 5 related questions you might explore:

1. How do you derive the quadratic formula for solving non-factorable quadratics?
2. What are the conditions for a system of equations to have no solutions, one solution, or infinite solutions?
3. Can a line intersect a parabola more than twice? Why or why not?
4. How would the solution change if the line had a different slope?
5. How do we verify solutions graphically in a system of nonlinear equations?

Tip: Always check your factored solution by expanding back to the original quadratic to avoid any errors.