To solve the system of equations given by
- \( y = x^2 + 48x + 31 \)
- \( y = 40x + 40 \)
we can set these two equations equal to each other:
\[ x^2 + 48x + 31 = 40x + 40 \]
Next, we can rearrange the equation to one side:
\[ x^2 + 48x + 31 - 40x - 40 = 0 \]
This simplifies to:
\[ x^2 + 8x - 9 = 0 \]
Now, we will factor the quadratic equation. We are looking for two numbers that multiply to \(-9\) and add to \(8\). The numbers are \(9\) and \(-1\). Therefore, we can factor the equation as:
\[ (x + 9)(x - 1) = 0 \]
Setting each factor to zero gives us:
\[ x + 9 = 0 \quad \Rightarrow \quad x = -9 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
Now we have two values for \(x\). Next, we find the corresponding \(y\) values for each \(x\) using the linear equation \(y = 40x + 40\).
- For \(x = -9\):
\[ y = 40(-9) + 40 = -360 + 40 = -320 \]
So one solution is \((-9, -320)\).
- For \(x = 1\):
\[ y = 40(1) + 40 = 40 + 40 = 80 \]
So the other solution is \((1, 80)\).
Thus, the coordinates of the solutions are:
\[ \boxed{(-9, -320)} \quad \text{and} \quad \boxed{(1, 80)} \]