Question

Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40 , y=−3x−5 ?(1 point)
Responses

(5,−10) and (−7,26)
left parenthesis 5 comma negative 10 right parenthesis and left parenthesis negative 7 comma 26 right parenthesis

(5,−10)
left parenthesis 5 comma negative 10 right parenthesis

(5,−20) and (−7,16)
left parenthesis negative 5 comma 20 right parenthesis and left parenthesis 7 comma negative 16 right parenthesis

(−7,26)

Answer 1

To solve the system of equations using the substitution method, we will substitute the linear equation \( y = -3x - 5 \) into the quadratic equation \( y = x^2 - x - 40 \).

1. Start with the equations:
\[
y = x^2 - x - 40
\]
\[
y = -3x - 5
\]

2. Substitute \( y \) in the quadratic equation:
\[
-3x - 5 = x^2 - x - 40
\]

3. Rearrange the equation to set it to zero:
\[
x^2 - x + 3x - 40 + 5 = 0
\]
\[
x^2 + 2x - 35 = 0
\]

4. Factor the quadratic equation:
\[
(x + 7)(x - 5) = 0
\]

5. Set each factor to zero:
\[
x + 7 = 0 \quad \Rightarrow \quad x = -7
\]
\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\]

6. Now, substitute these \( x \) values back into the linear equation \( y = -3x - 5 \) to find the corresponding \( y \) values.

For \( x = 5 \):
\[
y = -3(5) - 5 = -15 - 5 = -20
\]
So one solution is \( (5, -20) \).

For \( x = -7 \):
\[
y = -3(-7) - 5 = 21 - 5 = 16
\]
So the other solution is \( (-7, 16) \).

Thus, the solutions to the linear-quadratic system are \( (5, -20) \) and \( (-7, 16) \).

Based on the options you provided, no exact match was there, but it could be inferred you refer to:
\[
(5, -20) \text{ and } (-7, 16)
\]

Therefore, the solutions found are \( (5, -20) \) and \( (-7, 16) \).