To solve the linear-quadratic system given by the equations:
- \( y = x^2 - x - 40 \)
- \( y = -3x - 5 \)
you can use the substitution method.
Since both equations are equal to \( y \), we set them equal to each other:
\[ x^2 - x - 40 = -3x - 5 \]
Now, we rearrange the equation to set it to zero:
\[ x^2 - x + 3x - 40 + 5 = 0 \] \[ x^2 + 2x - 35 = 0 \]
Next, we can factor the quadratic equation:
\[ (x + 7)(x - 5) = 0 \]
Setting each factor to zero gives us the possible values for \( x \):
- \( x + 7 = 0 \) \(\Rightarrow x = -7\)
- \( x - 5 = 0 \) \(\Rightarrow x = 5\)
Now, we substitute these values back into one of the original equations to find the corresponding values of \( y \). We'll use the linear equation \( y = -3x - 5 \).
For \( x = -7 \):
\[ y = -3(-7) - 5 \] \[ y = 21 - 5 \] \[ y = 16 \] Thus, one solution is \((-7, 16)\).
For \( x = 5 \):
\[ y = -3(5) - 5 \] \[ y = -15 - 5 \] \[ y = -20 \] Thus, the other solution is \((5, -20)\).
Therefore, the solutions to the system are \((-7, 16)\) and \((5, -20)\).
Looking at your answer choices, it seems the correct response is:
\((-7, 16)\) and \((5, -20)\).