To solve the system of equations given by:

1. \( x + y = 18 \)
2. \( xy = 77 \)

We can start by solving for one variable in terms of the other using the first equation. We can express \( y \) in terms of \( x \):

\[
y = 18 - x
\]

Now we can substitute this expression for \( y \) into the second equation:

\[
x(18 - x) = 77
\]

Expanding this gives:

\[
18x - x^2 = 77
\]

Rearranging the equation leads to a standard quadratic equation:

\[
-x^2 + 18x - 77 = 0
\]

Multiplying the entire equation by -1 to make the leading coefficient positive:

\[
x^2 - 18x + 77 = 0
\]

Now we can solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -18 \), and \( c = 77 \). Plugging in these values:

\[
x = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 1 \cdot 77}}{2 \cdot 1}
\]

Calculating the discriminant:

\[
(-18)^2 = 324
\]
\[
4 \cdot 1 \cdot 77 = 308
\]
\[
b^2 - 4ac = 324 - 308 = 16
\]

Now substituting back into our formula for \( x \):

\[
x = \frac{18 \pm \sqrt{16}}{2}
\]
\[
x = \frac{18 \pm 4}{2}
\]

This gives us two possible values for \( x \):

1. \( x = \frac{22}{2} = 11 \)
2. \( x = \frac{14}{2} = 7 \)

Now, we can find the corresponding \( y \) values using \( y = 18 - x \):

1. If \( x = 11 \):
\[
y = 18 - 11 = 7
\]

2. If \( x = 7 \):
\[
y = 18 - 7 = 11
\]

Thus, the solutions to the system of equations are:

\[
(x, y) = (11, 7) \quad \text{and} \quad (7, 11)
\]