https://file.aiquickdraw.com/mathbot/file/1726158513767fnlwcpnf.jpg

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.

Solve the system of equations: y = 2x + 3 and y = x^2 + 3x - 3

Learn how to solve the system of equations with one linear and one quadratic equation: y = 2x + 3 and y = x^2 + 3x - 3. Includes step-by-step instructions and graphical interpretation for high school students.

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