To solve the linear-quadratic system using the elimination method, we have the equations:
- \( y = x^2 + 2x - 4 \)
- \( y = -x - 6 \)
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:
- \( x + 1 = 0 \) → \( x = -1 \)
- \( x + 2 = 0 \) → \( x = -2 \)
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:
\((-2, -4)\).