To solve the system of equations algebraically, we start with the two equations provided:
- \( y = x^2 - 2x - 1 \)
- \( y = 3x - 1 \)
Since both equations are equal to \( y \), we can set them equal to each other:
\[ x^2 - 2x - 1 = 3x - 1 \]
Next, we can move all terms to one side of the equation. First, we will subtract \( 3x \) and add \( 1 \) to both sides:
\[ x^2 - 2x - 1 - 3x + 1 = 0 \]
This simplifies to:
\[ x^2 - 5x + 0 = 0 \]
Factoring the equation:
\[ x(x - 5) = 0 \]
Setting each factor to zero gives us the solutions for \( x \):
- \( x = 0 \)
- \( x = 5 \)
Now, we will find the corresponding \( y \) values for each \( x \).
For \( x = 0 \):
Substituting into the second equation:
\[ y = 3(0) - 1 = -1 \]
So one solution is \( (0, -1) \).
For \( x = 5 \):
Substituting into the second equation:
\[ y = 3(5) - 1 = 15 - 1 = 14 \]
So the other solution is \( (5, 14) \).
Thus, the solutions to the system of equations are:
\[ (0, -1) \quad \text{and} \quad (5, 14) \]