Select the correct answer.

Solve the system of equations:
[tex]\[
\begin{array}{l}
y = x + 3 \\
y = x^2 - 2x - 1
\end{array}
\][/tex]

A. \((1, 4)\) and \((-4, -1)\)

B. \((-1, 4)\) and \((4, -1)\)

C. \((-1, 7)\) and \((4, 2)\)

D. [tex]\((-1, 2)\)[/tex] and [tex]\((4, 7)\)[/tex]



Answer :

To solve the system of equations:

[tex]\[ y = x + 3 \][/tex]
[tex]\[ y = x^2 - 2x - 1 \][/tex]

we need to find the points (x, y) that satisfy both equations simultaneously. Here are the detailed steps:

1. Set the Equations Equal to Each Other:

Since both expressions are equal to \( y \), we can set them equal to each other to find the values of \( x \):

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

2. Rewrite the Equation:

Rearrange the equation to set it to zero:

[tex]\[ x^2 - 2x - x - 1 - 3 = 0 \][/tex]

This simplifies to:

[tex]\[ x^2 - 3x - 4 = 0 \][/tex]

3. Solve the Quadratic Equation:

To solve the quadratic equation \( x^2 - 3x - 4 = 0 \), we can either factorize or use the quadratic formula. This particular quadratic can be factored:

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

Setting each factor to zero gives us the solutions:

[tex]\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \][/tex]
[tex]\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \][/tex]

4. Find Corresponding \( y \)-Values:

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

- For \( x = 4 \):
[tex]\[ y = 4 + 3 = 7 \][/tex]
So, one solution is \( (4, 7) \).

- For \( x = -1 \):
[tex]\[ y = -1 + 3 = 2 \][/tex]
So, another solution is \( (-1, 2) \).

5. List the Solutions:

The solutions to the system are the points where the lines intersect. These points are:

[tex]\[ (-1, 2) \][/tex]
[tex]\[ (4, 7) \][/tex]

6. Compare with Options:

Given the options:
- A: \((1, 4)\) and \((-4, -1)\)
- B: \((-1, 4)\) and \((4, -1)\)
- C: \((-1, 7)\) and \((4, 2)\)
- D: \((-1, 2)\) and \((4, 7)\)

We see that option D: \((-1, 2)\) and \((4, 7)\) matches our solutions perfectly.

Thus, the correct answer is:

[tex]\[ \boxed{D} \][/tex]