To solve the system of equations:
- \( y = x^2 + 13x + 13 \) (Equation 1)
- \( y = 2x - 17 \) (Equation 2)
We can set the two equations for \( y \) equal to each other since both are equal to \( y \):
\[ x^2 + 13x + 13 = 2x - 17 \]
Now, let's move all terms to one side of the equation:
\[ x^2 + 13x + 13 - 2x + 17 = 0 \]
This simplifies to:
\[ x^2 + 11x + 30 = 0 \]
Next, we will factor the quadratic equation. We need to find two numbers that multiply to \( 30 \) and add to \( 11 \). The numbers are \( 5 \) and \( 6 \).
Thus, we factor the equation as follows:
\[ (x + 5)(x + 6) = 0 \]
Setting each factor equal to zero gives:
\[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \] \[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \]
Now we have two \( x \) values: \( x = -5 \) and \( x = -6 \). We will now substitute these back into one of the original equations to find the corresponding \( y \) values.
Let's use Equation 2 (\( y = 2x - 17 \)).
For \( x = -5 \):
\[ y = 2(-5) - 17 = -10 - 17 = -27 \]
The coordinates are \( (-5, -27) \).
For \( x = -6 \):
\[ y = 2(-6) - 17 = -12 - 17 = -29 \]
The coordinates are \( (-6, -29) \).
Putting everything together, the system of equations has the following solutions:
\[ \text{Coordinates: } (-5, -27) \text{ and } (-6, -29) \]