To solve the given system of equations:
[tex]\[
\begin{array}{l}
x - y = 6 \\
y = x^2 - 6
\end{array}
\][/tex]
we start by expressing one variable in terms of the other using one of the equations. First, we solve the first equation for [tex]\(y\)[/tex]:
[tex]\[
x - y = 6 \implies y = x - 6.
\][/tex]
Next, we substitute [tex]\(y = x - 6\)[/tex] into the second equation:
[tex]\[
x - 6 = x^2 - 6.
\][/tex]
Now we solve for [tex]\(x\)[/tex] by simplifying and rearranging the equation:
[tex]\[
x - 6 = x^2 - 6 \implies x - 6 + 6 = x^2 - 6 + 6 \implies x = x^2.
\][/tex]
Rearrange it into a standard form quadratic equation:
[tex]\[
x^2 - x = 0.
\][/tex]
We can factor out an [tex]\(x\)[/tex]:
[tex]\[
x(x - 1) = 0.
\][/tex]
This gives us two solutions:
[tex]\[
x = 0 \quad \text{or} \quad x = 1.
\][/tex]
Now, substitute these values back into [tex]\(y = x - 6\)[/tex] to find the corresponding [tex]\(y\)[/tex]-values.
1. For [tex]\(x = 0\)[/tex]:
[tex]\[
y = 0 - 6 = -6.
\][/tex]
This gives the point [tex]\((0, -6)\)[/tex].
2. For [tex]\(x = 1\)[/tex]:
[tex]\[
y = 1 - 6 = -5.
\][/tex]
This gives the point [tex]\((1, -5)\)[/tex].
Thus, the solutions to the system of equations are [tex]\((0, -6)\)[/tex] and [tex]\((1, -5)\)[/tex].
From the given multiple-choice options, the correct answer is:
[tex]\[
\boxed{(0, -6) \text{ and } (1, -5)}
\][/tex]
