Question

Solve the linear-quadratic system by either evaluating the equations with the answer choices or using Geogebra graphing:

y=x2−x−6


y=x+2


(1 point)
Responses

(-2, 3)
(-2, 3)

(2, 4)
(2, 4)

(−2, 0) and (4, 6)
(−2, 0) and (4, 6)

(0, -2) and (6, 4)

Answer 1

To solve the system of equations given by the linear equation \(y = x + 2\) and the quadratic equation \(y = x^2 - x - 6\), we can set them equal to each other since they both equal \(y\):

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

Now, we can rearrange this equation to form a standard quadratic equation:

\[
x^2 - x - 6 - x - 2 = 0
\]
\[
x^2 - 2x - 8 = 0
\]

Next, we can factor the quadratic:

\[
(x - 4)(x + 2) = 0
\]

This gives us two solutions for \(x\):

\[
x - 4 = 0 \implies x = 4
\]
\[
x + 2 = 0 \implies x = -2
\]

Now, we can find the corresponding \(y\) values using the linear equation \(y = x + 2\):

1. For \(x = 4\):
\[
y = 4 + 2 = 6
\]
So one solution is \((4, 6)\).

2. For \(x = -2\):
\[
y = -2 + 2 = 0
\]
So the other solution is \((-2, 0)\).

Thus, the solutions to the system are:

\[
(-2, 0) \quad \text{and} \quad (4, 6)
\]

Among the options given, the correct response is:

**(−2, 0) and (4, 6)**.