To solve the system of equations
- \( y = x^2 + 32x + 44 \)
- \( y = 15x + 28 \)
we will set the two equations equal to each other:
\[ x^2 + 32x + 44 = 15x + 28 \]
Next, we will rearrange the equation by moving all terms to one side:
\[ x^2 + 32x + 44 - 15x - 28 = 0 \]
Simplifying this gives:
\[ x^2 + 17x + 16 = 0 \]
Now, we will factor the quadratic equation:
\[ (x + 1)(x + 16) = 0 \]
Setting each factor equal to zero gives us the solutions for \( x \):
- \( x + 1 = 0 \) → \( x = -1 \)
- \( x + 16 = 0 \) → \( x = -16 \)
Next, we substitute each \( x \) value back into one of the original equations to find the corresponding \( y \) values. We can use the linear equation \( y = 15x + 28 \).
- For \( x = -1 \):
\[ y = 15(-1) + 28 = -15 + 28 = 13 \]
Thus, one solution is \( (-1, 13) \).
- For \( x = -16 \):
\[ y = 15(-16) + 28 = -240 + 28 = -212 \]
Thus, the other solution is \( (-16, -212) \).
Now we have the solutions for the system of equations. The coordinates in exact form are:
\[ (-1, 13) \quad \text{and} \quad (-16, -212) \]