Question

Solve the system of equations.

y = x + 3
y = x2 - 2x - 1

A.
(-1, 2) and (4, 7)
B.
(-1, 7) and (4, 2)
C.
(-1, 4) and (4, -1)
D.
(1, 4) and (-4, -1)

Answer 1

To solve the system of equations given by:

1. \( y = x + 3 \)
2. \( y = x^2 - 2x - 1 \)

we can set the two equations equal to each other, as they both equal \( y \):

\[
x + 3 = x^2 - 2x - 1
\]

Now, we can rearrange this equation to set it to zero:

\[
0 = x^2 - 2x - 1 - x - 3
\]
\[
0 = x^2 - 3x - 4
\]

Now, we will factor the quadratic equation. We need two numbers that multiply to \(-4\) and add to \(-3\). The numbers that satisfy this are \(-4\) and \(1\):

\[
0 = (x - 4)(x + 1)
\]

Setting each factor equal to zero gives us the solutions for \(x\):

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

Now we will find the corresponding \(y\) values for both \(x\) values using the first equation \(y = x + 3\):

1. For \(x = 4\):

\[
y = 4 + 3 = 7
\]

So one solution is \((4, 7)\).

2. For \(x = -1\):

\[
y = -1 + 3 = 2
\]

So the other solution is \((-1, 2)\).

Thus, the solutions to the system of equations are the points \((-1, 2)\) and \((4, 7)\).

The correct answer is **A**: (-1, 2) and (4, 7).