To determine the number of real solutions for the system of equations given by \( y = -2x + 1 \) and \( y = -x^2 - 1 \), we can set the two expressions for \( y \) equal to each other:
\[ -2x + 1 = -x^2 - 1 \]
Now, rearranging this equation:
\[ -x^2 + 2x + 1 + 1 = 0 \] \[ -x^2 + 2x + 2 = 0 \]
Multiplying through by -1 to make the leading coefficient positive:
\[ x^2 - 2x - 2 = 0 \]
Now we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = -2 \):
First, calculate the discriminant (\( b^2 - 4ac \)):
\[ (-2)^2 - 4(1)(-2) = 4 + 8 = 12 \]
Since the discriminant is positive (\( 12 > 0 \)), this means there are two distinct real solutions for \( x \).
Thus, the system of equations has two real solutions.
Therefore, the correct response is:
The system of equations has two real solutions.
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