Question

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

(−7,26)
left parenthesis negative 7 comma 26 right parenthesis

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

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

(5,−20) and ,16

Answer 1

To solve the system of equations using the substitution method, we have:

1. \( y = x^2 - x - 40 \)
2. \( y = -3x - 5 \)

From the second equation, we can substitute \( y \) in the first equation:

\[
-3x - 5 = x^2 - x - 40
\]

Now, rewrite this equation to set it to 0:

\[
0 = x^2 - x - 40 + 3x + 5
\]

Combine like terms:

\[
0 = x^2 + 2x - 35
\]

Now, we can factor this quadratic equation:

\[
0 = (x + 7)(x - 5)
\]

Setting each factor to zero gives:

\[
x + 7 = 0 \quad \Rightarrow \quad x = -7
\]

\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\]

Now we have two \( x \) values: \( x = -7 \) and \( x = 5 \).

Next, we substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. Let's use \( y = -3x - 5 \).

1. For \( x = -7 \):

\[
y = -3(-7) - 5 = 21 - 5 = 16
\]
So, one solution is \( (-7, 16) \).

2. For \( x = 5 \):

\[
y = -3(5) - 5 = -15 - 5 = -20
\]
So, another solution is \( (5, -20) \).

Thus, the solutions to the linear-quadratic system are:

- \( (-7, 16) \)
- \( (5, -20) \)

Now, looking at your responses, the correct option is:

**(5,−20) and (−7,16)**