To solve the system of equations given by \( y = x^2 - 3x - 4 \) and \( y = -x - 1 \), we can set the equations equal to each other:
\[ x^2 - 3x - 4 = -x - 1 \]
Rearranging this gives:
\[ x^2 - 3x + x - 4 + 1 = 0 \] \[ x^2 - 2x - 3 = 0 \]
Next, we can factor the quadratic equation:
\[ (x - 3)(x + 1) = 0 \]
This gives us the solutions for \( x \):
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
Now we will find the corresponding \( y \) values for each \( x \).
-
For \( x = 3 \): \[ y = -3 - 1 = -4 \quad \Rightarrow \quad (3, -4) \]
-
For \( x = -1 \): \[ y = -(-1) - 1 = 0 \quad \Rightarrow \quad (-1, 0) \]
Thus, the solutions to the system of equations are \( (-1, 0) \) and \( (3, -4) \).
Checking the provided responses, the correct answer is:
(−1, 0) and (3, −4).