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 start with the two equations:

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

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

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

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

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

Simplifying this gives:

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

Now, we can factor the quadratic equation:

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

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

\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]

Now we have two values of \( x \): \( x = -1 \) and \( x = -2 \).

Next, we substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. We'll use the second equation \( y = -x - 6 \) for this.

1. For \( x = -1 \):
\[
y = -(-1) - 6 = 1 - 6 = -5
\]
So one solution is \( (-1, -5) \).

2. For \( x = -2 \):
\[
y = -(-2) - 6 = 2 - 6 = -4
\]
So the other solution is \( (-2, -4) \).

Thus, the full solutions to the system are:

\[
(-1, -5) \quad \text{and} \quad (-2, -4)
\]

The answer is \( (-2, -4) \).