No Real Solutions Quick Check

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.**