To solve the system of equations

\[
\begin{cases}
2x + y = 4 \\
xy = 1
\end{cases}
\]

we can use substitution. From the first equation, we can express \(y\) in terms of \(x\):

\[
y = 4 - 2x
\]

Now, substitute \(y\) into the second equation:

\[
x(4 - 2x) = 1
\]

Expanding this gives:

\[
4x - 2x^2 = 1
\]

Rearranging the equation:

\[
2x^2 - 4x + 1 = 0
\]

Now we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -4\), and \(c = 1\).

Calculating the discriminant:

\[
b^2 - 4ac = (-4)^2 - 4 \cdot 2 \cdot 1 = 16 - 8 = 8
\]

Now substituting into the formula:

\[
x = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}
\]

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

1. \( x_1 = 1 + \frac{\sqrt{2}}{2} \)
2. \( x_2 = 1 - \frac{\sqrt{2}}{2} \)

Next, we can find the corresponding \(y\) values using \(y = 4 - 2x\):

For \(x_1\):

\[
y_1 = 4 - 2 \left(1 + \frac{\sqrt{2}}{2}\right) = 4 - 2 - \sqrt{2} = 2 - \sqrt{2}
\]

For \(x_2\):

\[
y_2 = 4 - 2 \left(1 - \frac{\sqrt{2}}{2}\right) = 4 - 2 + \sqrt{2} = 2 + \sqrt{2}
\]

Thus, the solutions are:

\[
\left(1 + \frac{\sqrt{2}}{2}, 2 - \sqrt{2}\right) \quad \text{and} \quad \left(1 - \frac{\sqrt{2}}{2}, 2 + \sqrt{2}\right)
\]