To determine the number of real solutions for the system of equations \( y = -2x + 1 \) and \( y = -x^2 - 1 \), we can set the two equations equal to each other since they both equal \( y \):
\[ -2x + 1 = -x^2 - 1 \]
Now, rearranging this equation gives:
\[ x^2 - 2x + 2 = 0 \]
Next, we can use the quadratic formula to find the solutions, where \( a = 1, b = -2, c = 2 \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \] \[ x = \frac{2 \pm \sqrt{4 - 8}}{2} \] \[ x = \frac{2 \pm \sqrt{-4}}{2} \] \[ x = \frac{2 \pm 2i}{2} \] \[ x = 1 \pm i \]
Since the discriminant \( b^2 - 4ac \) is negative (\( -4 \)), this indicates that there are no real solutions to the quadratic equation.
Thus, the best characterization for the number of real solutions for the system of equations is:
The system of equations has no real solutions.
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