Use the elimination method to solve the linear-quadratic system: y = x^2 + 10, y = -7x - 2

To solve the system of equations using the elimination method, we'll set the two equations equal to each other since both are equal to \( y \).

The two equations are:

1. \( y = x^2 + 10 \)
2. \( y = -7x - 2 \)

Setting them equal to each other gives:

\[
x^2 + 10 = -7x - 2
\]

To solve for \( x \), we can rearrange the equation by moving all terms to one side:

\[
x^2 + 7x + 10 + 2 = 0
\]

This simplifies to:

\[
x^2 + 7x + 12 = 0
\]

Next, we will factor the quadratic equation:

\[
(x + 3)(x + 4) = 0
\]

Setting each factor to zero gives us the solutions for \( x \):

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x + 4 = 0 \) → \( x = -4 \)

Now that we have the \( x \)-values, we can substitute these back into either original equation to find the corresponding \( y \)-values. We'll use the second equation, \( y = -7x - 2 \).

**For \( x = -3 \):**

\[
y = -7(-3) - 2 = 21 - 2 = 19
\]

The first solution is \( (-3, 19) \).

**For \( x = -4 \):**

\[
y = -7(-4) - 2 = 28 - 2 = 26
\]

The second solution is \( (-4, 26) \).

Putting it all together, the solutions to the system of equations are:

\[
\boxed{(-3, 19)} \quad \text{and} \quad \boxed{(-4, 26)}
\]