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 \( y \):
\[ -2x + 1 = x^2 - 1 \]
Rearranging this equation gives us:
\[ x^2 + 2x - 2 = 0 \]
Now, we can use the quadratic formula to find the roots of this quadratic equation:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2 \), and \( c = -2 \). Substituting these values into the formula:
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]
\[ x = \frac{-2 \pm \sqrt{4 + 8}}{2} \]
\[ x = \frac{-2 \pm \sqrt{12}}{2} \]
\[ x = \frac{-2 \pm 2\sqrt{3}}{2} \]
\[ x = -1 \pm \sqrt{3} \]
This gives us two distinct real solutions for \( x \):
\[ x = -1 + \sqrt{3} \quad \text{and} \quad x = -1 - \sqrt{3} \]
Since there are two distinct values for \( x \), we can conclude that the system of equations has two real solutions.
Thus, the best characterization of the number of real solutions for the system of equations is:
The system of equations has two real solutions.