To solve the system of equations given by:
- \( y = x^2 + 38x + 37 \) (Equation 1)
- \( y = 50x + 1 \) (Equation 2)
We can set the two equations equal to each other since they both equal \( y \):
\[ x^2 + 38x + 37 = 50x + 1 \]
Now, we'll rearrange the equation to gather all terms on one side:
\[ x^2 + 38x + 37 - 50x - 1 = 0 \]
Combining like terms gives us:
\[ x^2 - 12x + 36 = 0 \]
Next, we can factor the quadratic equation:
\[ (x - 6)(x - 6) = 0 \]
Or simply:
\[ (x - 6)^2 = 0 \]
This implies that:
\[ x - 6 = 0 \implies x = 6 \]
Now, to find the corresponding \( y \) value, we can substitute \( x = 6 \) back into either of the original equations. It's often easiest to use the linear equation (Equation 2):
\[ y = 50(6) + 1 = 300 + 1 = 301 \]
Thus, the solution to the system of equations, or the coordinates of the intersection point, is:
\[ \boxed{(6, 301)} \]