Question

Solve the system by substitution.

y=−x^2+2x+10

y=x+4

(1 point)
Responses

(−1,3),(2,6)
open paren negative 1 comma 3 close paren comma open paren 2 comma 6 close paren

(0,4),(4,8)
open paren 0 comma 4 close paren comma open paren 4 comma 8 close paren

(1,5),(5,9)
open paren 1 comma 5 close paren comma open paren 5 comma 9 close paren

(3,7),(−2,2)
open paren 3 comma 7 close paren comma open paren negative 2 comma 2 close paren
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Answer 1

To solve the system of equations by substitution, we will set the two equations equal to each other since both are equal to \( y \).

1. **Given equations:**
\[
y = -x^2 + 2x + 10 \quad \text{(1)}
\]
\[
y = x + 4 \quad \text{(2)}
\]

2. **Substitute equation (2) into equation (1):**
\[
x + 4 = -x^2 + 2x + 10
\]

3. **Rearranging the equation:**
\[
0 = -x^2 + 2x - x + 10 - 4
\]
\[
0 = -x^2 + x + 6
\]
\[
x^2 - x - 6 = 0 \quad \text{(multiply the entire equation by -1)}
\]

4. **Factoring the quadratic:**
\[
(x - 3)(x + 2) = 0
\]

5. **Setting each factor to zero:**
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]

6. **Now, find corresponding \( y \) values for each \( x \):**

- For \( x = 3 \):
\[
y = 3 + 4 = 7
\]
So one solution is \( (3, 7) \).

- For \( x = -2 \):
\[
y = -2 + 4 = 2
\]
So the second solution is \( (-2, 2) \).

7. **Final solutions:**
The solutions to the system are \( (3, 7) \) and \( (-2, 2) \).

Thus, the answer is:
\[
\text{(3,7),(−2,2)}
\]