Answer:
Option C. The given system has two solutions.
Solution:
The given equations are,
![y = -x + 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cq9v0mx436u4o1oyntjro5fp2jaut2y3f3.png)
![y = -x^2 + 4x-2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ywkpchonosmo5q2b0xofl7lzru2ny2dcwt.png)
From the equation we can say,
![-x+1 = -x^2+ 4x-2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jdq7pmv74t63f3wmy2n8lc2wagcnmpw3ha.png)
![\Rightarrow-x^(2)+4 x-2+x-1=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/p34vpmzg5z0xzdqpnaoxt0yct0z7otwrfq.png)
![\Rightarrow-x^(2)+5 x-3=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rkppqrii0otk0w0ffd1f4d18qrm7gvh7hh.png)
We know that the quadratic formula to solve this,
x has two values which are
Here, a = (-1), b = 5 , c = -3
So,
![x=\frac{(-5+\sqrt{(5)^(2)-4 x(-1)} *(-3))}{2 *(-1)}=((-5+√(25-12)))/(-2)=((-5+√(13)))/(-2)=((5-√(13)))/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vdd01ps1n6dym4s143mz77oy6puois0ug4.png)
Again
![x=((5+√(13)))/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/r29bc7unqbo2j50whyargn021ssvs86ejj.png)
Hence, x has two solutions.