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 (
).
Remaining Attempts : 3

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. \( y = -x - 6 \)

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

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

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

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

Combining like terms, we get:

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

Next, we can factor this quadratic equation:

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

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

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

Now we'll find the corresponding \( y \) values for each \( x \) using one of the original equations. We'll use \( y = -x - 6 \):

For \( x = -1 \):
\[
y = -(-1) - 6 = 1 - 6 = -5
\]
So, we have the solution \( (-1, -5) \).

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

Thus, the solutions to the system are:

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

The final answer is:

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