Question

Solve the system of equations:
y = 2x
y= x2 - 8

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

O (-4, 8) and (2, - 4)

O (-2, -4) and (4, 8)

0 (-4, -8) and (2, 4)

Answer 1

To solve the system of equations given by:

1. \( y = 2x \)
2. \( y = x^2 - 8 \)

We can set the two equations equal to each other since they both equal \( y \):

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

Rearranging this equation gives us:

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

Now, we can factor this quadratic equation. We look for two numbers that multiply to \(-8\) and add to \(-2\). Those numbers are \(-4\) and \(2\). Thus, we can factor the equation as:

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

Setting each factor equal to zero gives us the \(x\)-values:

1. \( x - 4 = 0 \) → \( x = 4 \)
2. \( x + 2 = 0 \) → \( x = -2 \)

Now we can find the corresponding \(y\)-values for each \(x\):

1. For \( x = 4 \):

\[
y = 2(4) = 8
\]

So one solution is \( (4, 8) \).

2. For \( x = -2 \):

\[
y = 2(-2) = -4
\]

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

The solutions to the system of equations are:

\[
(-2, -4) \quad \text{and} \quad (4, 8)
\]

Therefore, the correct answer is:

\[
(-2, -4) \text{ and } (4, 8)
\]