https://file.aiquickdraw.com/mathbot/file/1726618558434fzufingr.jpg

The system of equations provided in the image is:

\[
\begin{cases}
y = x^2 + 2x + 5 \\
y = 5x + 15
\end{cases}
\]

To solve this system, we can use substitution since both equations are equal to \( y \).

### Step 1: Set the two equations equal to each other:
\[
x^2 + 2x + 5 = 5x + 15
\]

### Step 2: Rearrange the equation to set it equal to zero:
\[
x^2 + 2x + 5 - 5x - 15 = 0
\]
\[
x^2 - 3x - 10 = 0
\]

### Step 3: Solve the quadratic equation:
We can factor the quadratic:
\[
(x - 5)(x + 2) = 0
\]

This gives the solutions:
\[
x = 5 \quad \text{or} \quad x = -2
\]

### Step 4: Substitute these \( x \)-values into one of the original equations to find \( y \):
- When \( x = 5 \):
\[
y = 5(5) + 15 = 25 + 15 = 40
\]
So, one solution is \( (5, 40) \).

- When \( x = -2 \):
\[
y = 5(-2) + 15 = -10 + 15 = 5
\]
So, the other solution is \( (-2, 5) \).

### Final Answer:
The solutions are \( (5, 40) \) and \( (-2, 5) \).

Let me know if you would like more details on any step! Here are a few related questions to explore:

1. How do you determine whether a system of equations has one, two, or infinite solutions?
2. What is the geometric interpretation of the solutions to this system?
3. How do you factor a quadratic equation if it can't be easily factored like this one?
4. Can you solve the same system using the method of substitution or elimination?
5. What would the solution be if the quadratic had no real roots?

**Tip:** Always double-check the discriminant in a quadratic equation to see if real solutions exist before factoring.

Solve the system of equations: y = x^2 + 2x + 5 and y = 5x + 15.

Learn how to solve a system of equations consisting of y = x^2 + 2x + 5 and y = 5x + 15 using substitution and factoring techniques. This problem involves algebraic concepts such as quadratic equations and factoring, suitable for students in Grades 8-10.

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