To determine one of the solutions to the given system of equations:
[tex]\[
\left\{\begin{array}{r}
y - 3 = x \\
x^2 - 6x + 13 = y
\end{array}\right.
\][/tex]
we will follow these steps for each of the given points: [tex]\((-5,2)\)[/tex], [tex]\((-2,1)\)[/tex], [tex]\((2,5)\)[/tex], and [tex]\((8,5)\)[/tex]:
### Checking Point [tex]\((-5, 2)\)[/tex]:
1. Substitute [tex]\((-5, 2)\)[/tex] into the first equation [tex]\(y - 3 = x\)[/tex]:
[tex]\[2 - 3 = -1 \neq -5\][/tex]
Since this does not hold, [tex]\((-5, 2)\)[/tex] is not a solution.
### Checking Point [tex]\((-2, 1)\)[/tex]:
1. Substitute [tex]\((-2, 1)\)[/tex] into the first equation [tex]\(y - 3 = x\)[/tex]:
[tex]\[1 - 3 = -2 \][/tex]
This holds true.
2. Now, substitute [tex]\((-2, 1)\)[/tex] into the second equation [tex]\(x^2 - 6x + 13 = y\)[/tex]:
[tex]\[ (-2)^2 - 6(-2) + 13 = 4 + 12 + 13 = 29 \neq 1\][/tex]
Since the second equation does not hold, [tex]\((-2, 1)\)[/tex] is not a solution.
### Checking Point [tex]\((2, 5)\)[/tex]:
1. Substitute [tex]\((2, 5)\)[/tex] into the first equation [tex]\(y - 3 = x\)[/tex]:
[tex]\[5 - 3 = 2 \][/tex]
This holds true.
2. Now, substitute [tex]\((2, 5)\)[/tex] into the second equation [tex]\(x^2 - 6x + 13 = y\)[/tex]:
[tex]\[ (2)^2 - 6(2) + 13 = 4 - 12 + 13 = 5\][/tex]
This holds true.
Since both equations are satisfied, [tex]\((2, 5)\)[/tex] is a solution to the system.
### Checking Point [tex]\((8, 5)\)[/tex]:
1. Substitute [tex]\((8, 5)\)[/tex] into the first equation [tex]\(y - 3 = x\)[/tex]:
[tex]\[5 - 3 = 2 \neq 8\][/tex]
Since this does not hold, [tex]\((8, 5)\)[/tex] is not a solution.
Therefore, after checking all points, one of the solutions to the system of equations is:
[tex]\[
(2, 5)
\][/tex]
