To solve the system of equations given by
- \( y = 3x + 25 \)
- \( y = 13x^2 + 3x + 12 \)
we can set the two expressions for \( y \) equal to each other:
\[ 3x + 25 = 13x^2 + 3x + 12 \]
Next, we will simplify this equation by subtracting \( 3x + 25 \) from both sides:
\[ 0 = 13x^2 + 3x + 12 - 3x - 25 \]
This simplifies to:
\[ 0 = 13x^2 - 13 \]
Now, we can factor out the common term:
\[ 0 = 13(x^2 - 1) \]
Next, we set each factor equal to zero:
\[ 13 = 0 \quad \text{(not applicable)} \] \[ x^2 - 1 = 0 \]
We solve \( x^2 - 1 = 0 \):
\[ x^2 = 1 \]
Taking the square root of both sides, we find:
\[ x = 1 \quad \text{or} \quad x = -1 \]
Now we will find the corresponding \( y \)-coordinates for each \( x \) value using the first equation:
- For \( x = 1 \):
\[ y = 3(1) + 25 = 3 + 25 = 28 \]
- For \( x = -1 \):
\[ y = 3(-1) + 25 = -3 + 25 = 22 \]
Thus, the coordinates of the solutions are:
\[ (1, 28) \quad \text{and} \quad (-1, 22) \]
Therefore, the solutions to the system of equations are:
\[ \boxed{(1, 28)} \quad \text{and} \quad \boxed{(-1, 22)} \]