Use the substitution method to find the solutions to the linear-quadratic

To solve the system of equations using the substitution method, we have the two equations:

1. \( y = x^2 - x - 10 \)
2. \( y = -3x - 5 \)

Since both expressions are equal to \( y \), we can set them equal to each other:

\[
x^2 - x - 10 = -3x - 5
\]

Now, we will move all terms to one side of the equation:

\[
x^2 - x + 3x - 10 + 5 = 0
\]

This simplifies to:

\[
x^2 + 2x - 5 = 0
\]

Next, we will factor this equation. However, since it does not factor nicely, we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 2 \), and \( c = -5 \).

Calculating the discriminant:

\[
b^2 - 4ac = 2^2 - 4(1)(-5) = 4 + 20 = 24
\]

Now, substituting back into the formula:

\[
x = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6}
\]

Thus, we have two values for \( x \):

1. \( x = -1 + \sqrt{6} \)
2. \( x = -1 - \sqrt{6} \)

Now we need to find the corresponding \( y \) values using one of the original equations (let’s use the second equation):

For \( x = -1 + \sqrt{6} \):

\[
y = -3(-1 + \sqrt{6}) - 5 = 3 - 3\sqrt{6} - 5 = -2 - 3\sqrt{6}
\]

For \( x = -1 - \sqrt{6} \):

\[
y = -3(-1 - \sqrt{6}) - 5 = 3 + 3\sqrt{6} - 5 = -2 + 3\sqrt{6}
\]

Thus, the solutions to the system of equations are:

1. \( (-1 + \sqrt{6}, -2 - 3\sqrt{6}) \)
2. \( (-1 - \sqrt{6}, -2 + 3\sqrt{6}) \)

Now, let's compare our solutions to the provided options:
- (5, -10) and (-7, 26)
- (5, -10)
- (-7, 26)
- (5, -20) and (-7, 16)

Based on our calculations, it appears our precise solutions do not match the given options directly (as they involve square roots and do not yield integer solutions).

If you have any specific options listed below that you would like me to analyze further based on approximation or verification against those solutions, please let me know!