Question

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

Answer 1

To solve the system using the elimination method, we want to eliminate one variable by adding or subtracting the two equations.

First, let's rewrite the equations in standard form:

y = x^2 + 2x - 4 ---> x^2 + 2x - 4 - y = 0 ---> (1)
y = -x - 6 ---> x + y = -6 ---> x + y + 6 = 0 ---> (2)

Now, let's subtract equation (2) from equation (1):

(x^2 + 2x - 4 - y) - (x + y + 6) = 0

Simplifying, we get:

x^2 + 2x - 4 - x - y - 6 = 0
x^2 + x - 10 = 0

Now, we have a quadratic equation in terms of x. Let's solve this equation by factoring or by using the quadratic formula:

(x + 5)(x - 2) = 0

Setting each factor equal to zero and solving for x, we have:

x + 5 = 0 --> x = -5
x - 2 = 0 --> x = 2

Now, substitute these values of x back into the original equations to find the corresponding values of y.

For x = -5:
y = -x - 6
y = -(-5) - 6
y = 5 - 6
y = -1

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

For x = 2:
y = -x - 6
y = -(2) - 6
y = -2 - 6
y = -8

So, another solution is (2, -8).

Therefore, the solutions to the linear-quadratic system are (-1, -5) and (2, -8).