To solve the linear-quadratic system using the substitution method, we have the following equations:
- \( y = x^2 - x - 40 \) (quadratic)
- \( y = -3x - 5 \) (linear)
We can substitute the expression for \( y \) from the linear equation into the quadratic equation:
\[ -3x - 5 = x^2 - x - 40 \]
Now, rearranging the equation gives us:
\[ x^2 - x - 40 + 3x + 5 = 0 \]
Combining like terms:
\[ x^2 + 2x - 35 = 0 \]
Next, we will factor the quadratic:
\[ (x + 7)(x - 5) = 0 \]
Setting each factor to zero gives us the solutions for \( x \):
- \( x + 7 = 0 \Rightarrow x = -7 \)
- \( x - 5 = 0 \Rightarrow x = 5 \)
Now, we will find the corresponding \( y \) values for each \( x \) using the linear equation \( y = -3x - 5 \).
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, the other solution is \( (5, -20) \).
Thus, the solutions to the system are \( (5, -20) \) and \( (-7, 16) \).
The choices given were:
- (5, -20) and (-7, 16)
So the answer is: (5, -20) and (-7, 16).