Use the elimination method to solve the linear-quadratic system y=x^2+2x−4, y=−x−6.

Answer 1

To solve the system using the elimination method, we need to eliminate one variable from the equations.

Let's start by multiplying the second equation by 2, so that the coefficients of y in both equations are the same:

2(y) = 2(-x - 6)
2y = -2x - 12

Now we have the system:

y = x^2 + 2x - 4
2y = -2x - 12

Since the coefficient of y in the second equation is 2, we can multiply the first equation by 2 as well:

2(y) = 2(x^2 + 2x - 4)
2y = 2x^2 + 4x - 8

Now we have the system:

2y = 2x^2 + 4x - 8
2y = -2x - 12

We can subtract the second equation from the first to eliminate y:

2x^2 + 4x - 8 - (-2x - 12) = 0

Simplifying, we get:

2x^2 + 4x - 8 + 2x + 12 = 0
2x^2 + 6x + 4 = 0

Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula. In this case, the equation factors as follows:

2(x + 1)(x + 2) = 0

Setting each factor equal to zero, we have:

x + 1 = 0 or x + 2 = 0

Solving each equation, we find:

x = -1 or x = -2

Now we substitute these values back into one of the original equations to find the corresponding values of y.

Using the equation y = x^2 + 2x - 4:

For x = -1:
y = (-1)^2 + 2(-1) - 4
y = 1 - 2 - 4
y = -5

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

For x = -2:
y = (-2)^2 + 2(-2) - 4
y = 4 - 4 - 4
y = -4

So the other solution is (-2, -4).

Therefore, the system has two solutions: (-1, -5) and (-2, -4).