Question

Use the elimination method to solve the linear-quadratic system y=x^2+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 need to eliminate one variable by adding or subtracting the equations. In this case, we can eliminate the variable y by adding the two equations together.

We have:
y = x^2 + 2x - 4
y = -x - 6

Adding the equations:
(x^2 + 2x - 4) + (-x - 6) = 0

Simplifying:
x^2 + x - 10 = 0

Now we can solve this quadratic equation by factoring or using the quadratic formula. Factoring gives us:
(x + 5)(x - 2) = 0

Setting each factor equal to zero:
x + 5 = 0 or x - 2 = 0

Solving for x:
x = -5 or x = 2

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

For x = -5:
y = (-5)^2 + 2(-5) - 4
y = 25 - 10 - 4
y = 11

So one solution is (-5, 11).

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

So the other solution is (2, 4).

Therefore, the solutions to the linear-quadratic system are (-5, 11) and (2, 4).