To solve the system of equations using the substitution method, we have the following two equations:
- \( y = x^2 - x - 40 \) (equation 1)
- \( y = -3x - 5 \) (equation 2)
Step 1: Set the two equations equal to each other:
\[ x^2 - x - 40 = -3x - 5 \]
Step 2: Rearrange the equation to set it to zero:
\[ x^2 - x + 3x - 40 + 5 = 0 \] \[ x^2 + 2x - 35 = 0 \]
Step 3: Factor the quadratic equation:
\[ (x + 7)(x - 5) = 0 \]
Step 4: Set each factor equal to zero and solve for \( x \):
- \( x + 7 = 0 \) → \( x = -7 \)
- \( x - 5 = 0 \) → \( x = 5 \)
Step 5: Now substitute each value of \( x \) back into one of the original equations to find the corresponding \( y \) values. We'll use equation 2 (\( y = -3x - 5 \)).
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 system are:
- \( (5, -20) \)
- \( (-7, 16) \)
Thus, the correct response is:
(5,−20) and (−7,16).