Answer:
Option C. The given system has two solutions.
Solution:
The given equations are,
[tex]y = -x + 1[/tex]
[tex]y = -x^2 + 4x-2[/tex]
From the equation we can say,
[tex]-x+1 = -x^2+ 4x-2[/tex]
[tex]\Rightarrow-x^{2}+4 x-2+x-1=0[/tex]
[tex]\Rightarrow-x^{2}+5 x-3=0[/tex]
We know that the quadratic formula to solve this,
x has two values which are [tex]\frac{(-b+\sqrt{b^{2}-4 a c})}{2 a} \ and \ \frac{(-b-\sqrt{\left.b^{2}-4 a c\right)}}{2 a}[/tex]
Here, a = (-1), b = 5 , c = -3
So, [tex]x=\frac{(-5+\sqrt{(5)^{2}-4 x(-1)} \times(-3))}{2 \times(-1)}=\frac{(-5+\sqrt{25-12})}{-2}=\frac{(-5+\sqrt{13})}{-2}=\frac{(5-\sqrt{13})}{2}[/tex]
Again [tex]x=\frac{(5+\sqrt{13})}{2}[/tex]
Hence, x has two solutions.
