Question

Solve the following system of equations algebraically:
y, equals, x, squared, minus, x, minus, 65
y=
x
2
−x−65
y, equals, minus, x, minus, 1
y=
−x−1

Answer 1

To solve the system of equations given by:

1. \( y = x^2 - x - 65 \)
2. \( y = -x - 1 \)

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

\[
x^2 - x - 65 = -x - 1
\]

Next, let's simplify this equation. We'll start by adding \( x + 1 \) to both sides:

\[
x^2 - x - 65 + x + 1 = 0
\]

This simplifies to:

\[
x^2 - 64 = 0
\]

Now, we can factor the equation:

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

Setting each factor to zero, we have:

1. \( x - 8 = 0 \) ⟹ \( x = 8 \)
2. \( x + 8 = 0 \) ⟹ \( x = -8 \)

Now we will find the corresponding \( y \) values for each \( x \):

**For \( x = 8 \):**

Using the second equation \( y = -x - 1 \):

\[
y = -8 - 1 = -9
\]

This gives the first solution:

\[
(8, -9)
\]

**For \( x = -8 \):**

Again, using the second equation \( y = -x - 1 \):

\[
y = -(-8) - 1 = 8 - 1 = 7
\]

This gives the second solution:

\[
(-8, 7)
\]

Therefore, the solution set for the system of equations is:

\[
\boxed{(8, -9) \text{ and } (-8, 7)}
\]