To solve the system of equations using the substitution method, we have:
- \( y = x^2 - x - 40 \)
- \( y = -3x - 5 \)
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 \( y = -3x - 5 \).
- For \( x = -7 \):
\[ y = -3(-7) - 5 = 21 - 5 = 16 \] So, one solution is \( (-7, 16) \).
- 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)