Question

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 system of equations using the elimination method, we have the following two equations:

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

We'll set these equations equal to each other since both are equal to \( y \):

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

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

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

This simplifies to:

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

Next, we factor the quadratic equation:

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

Setting each factor equal to zero gives us two possible solutions for \( x \):

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

Now, we can find the corresponding \( y \) values by substituting these \( x \) values back into either of the original equations. We'll use the second equation \( y = -x - 6 \).

For \( x = -1 \):

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

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

For \( x = -2 \):

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

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

Thus, the two solutions to the system of equations are:

1. \((-1, -5)\)
2. \((-2, -4)\)

The answer you provided already contains one solution, so the full pair of solutions is:

The solutions are \((-1, -5)\) and \((-2, -4)\).