Question

Use the elimination method to solve the linear-quadratic system y=x2+2x−4
, y=−x−6
.(1 point)
The solutions are (−1,−5)
and (
).

Answer 1

To solve the linear-quadratic system using the elimination method, we have the equations:

1. \( y = x^2 + 2x - 4 \)
2. \( y = -x - 6 \)

We will equate the two expressions for \( y \):

\[
x^2 + 2x - 4 = -x - 6
\]

Next, we will rearrange the equation and set it to 0:

\[
x^2 + 2x + x - 4 + 6 = 0
\]

This simplifies to:

\[
x^2 + 3x + 2 = 0
\]

Now, we can factor the quadratic equation:

\[
(x + 1)(x + 2) = 0
\]

Setting each factor equal to zero gives us:

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

Now we will find the corresponding \( y \) values for these \( x \) values using the linear equation \( y = -x - 6 \).

For \( x = -1 \):

\[
y = -(-1) - 6 = 1 - 6 = -5
\]

So one solution is \((-1, -5)\).

For \( x = -2 \):

\[
y = -(-2) - 6 = 2 - 6 = -4
\]

Thus, the second solution is \((-2, -4)\).

In conclusion, the solutions to the system are \((-1, -5)\) and \((-2, -4)\). The complete answer is:

\((-2, -4)\).