Use the elimination method to solve the following linear-quadratic system: Y - x^2 = x - 42 & x + y = 6

To solve the system of equations using the elimination method, we first rewrite both equations:

1. \( y - x^2 = x - 42 \) becomes \( y = x^2 + x - 42 \)
2. \( x + y = 6 \) can be rewritten as \( y = 6 - x \)

Now we can set the two expressions for \( y \) equal to each other:

\[
x^2 + x - 42 = 6 - x
\]

Next, we simplify this equation:

\[
x^2 + x + x - 42 - 6 = 0
\]

\[
x^2 + 2x - 48 = 0
\]

Now, we can factor the quadratic:

\[
(x + 8)(x - 6) = 0
\]

This gives us:

\[
x + 8 = 0 \quad \text{or} \quad x - 6 = 0
\]

Which leads to:

\[
x = -8 \quad \text{or} \quad x = 6
\]

Now we use these values of \( x \) to find the corresponding values of \( y \):

1. For \( x = -8 \):
\[
y = 6 - (-8) = 6 + 8 = 14
\]

2. For \( x = 6 \):
\[
y = 6 - 6 = 0
\]

Thus, the solutions to the system are:
- \( (-8, 14) \)
- \( (6, 0) \)

Looking at the provided answer choices, the correct answer is:

**1. (-8, 14) and (6, 0)**