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]
