what is one of the solutions to the following system?\n$\begin{cases}y - 3=x\\x^{2}-6x + 13=yend{cases}$\n$(…

Explanation:

Step1: Substitute x in the second - equation

Given $y - 3=x$, substitute $x$ into $x^{2}-6x + 13=y$. We get $(y - 3)^{2}-6(y - 3)+13=y$. Expand: $y^{2}-6y + 9-6y+18 + 13=y$. Simplify to $y^{2}-12y+40=y$, then $y^{2}-13y + 40=0$.

Step2: Solve the quadratic equation for y

Factor the quadratic equation $y^{2}-13y + 40=0$. We have $(y - 5)(y - 8)=0$. So $y=5$ or $y = 8$.

Step3: Find the corresponding x - values

When $y = 5$, since $y-3=x$, then $x=y - 3=5 - 3=2$. When $y = 8$, $x=y - 3=8 - 3=5$. The solution $(x,y)=(2,5)$ is one of the solutions.

Answer:

C. (2, 5)