Solve the system algebraically.    

y= -x2 + 9

y= x2 + 1

Group of answer choices

(3, 1) and (1, 3)

(-2, 5) and (2, 5)

(-2, 2) and (5, 5)

(5, -2) and (5, 2)

To solve the system algebraically, we'll set the two equations equal to each other since they both equal \( y \):

### Given equations:
1. \( y = -x^2 + 9 \)
2. \( y = x^2 + 1 \)

### Step 1: Set the equations equal to each other
\[
-x^2 + 9 = x^2 + 1
\]

### Step 2: Combine like terms
\[
9 - 1 = x^2 + x^2
\]
\[
8 = 2x^2
\]

### Step 3: Solve for \( x^2 \)
\[
x^2 = \frac{8}{2} = 4
\]

### Step 4: Solve for \( x \)
\[
x = \pm \sqrt{4} = \pm 2
\]

### Step 5: Solve for \( y \)
Substitute \( x = 2 \) and \( x = -2 \) into either equation. Using \( y = x^2 + 1 \):

- When \( x = 2 \):
\[
y = 2^2 + 1 = 4 + 1 = 5
\]

- When \( x = -2 \):
\[
y = (-2)^2 + 1 = 4 + 1 = 5
\]

### Step 6: Write the solution pairs
The solutions are:
\[
(-2, 5) \quad \text{and} \quad (2, 5)
\]

### Final Answer:
\[
\boxed{(-2, 5) \text{ and } (2, 5)}
\]

Solve the system algebraically. y = -x^2 + 9, y = x^2 + 1

Learn how to solve the system of equations y = -x^2 + 9 and y = x^2 + 1 algebraically with a clear step-by-step guide. This solution involves setting equations equal, solving quadratic equations, and identifying the solution pairs. Ideal for students in Grades 9-10.

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