To analyze 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 the equation gives:
\[ -x^2 + 2x + 1 + 1 = 0 \]
This simplifies to:
\[ -x^2 + 2x + 2 = 0 \]
Multiplying through by -1 to make the quadratic standard gives:
\[ x^2 - 2x - 2 = 0 \]
Next, we can use the discriminant to determine the number of real solutions. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
For our equation \( x^2 - 2x - 2 = 0 \):
- \( a = 1 \)
- \( b = -2 \)
- \( c = -2 \)
Calculating the discriminant:
\[ D = (-2)^2 - 4(1)(-2) = 4 + 8 = 12 \]
Since the discriminant \( D = 12 \) is positive, this indicates that there are two distinct real solutions for the equation.
Thus, the best characterization of the number of real solutions for the system of equations is:
The system of equations has two real solutions.