Example 8 Given $x+y=2, x^{2}+y^{2}=\frac{5}{2}$. Then $x^{4}+y^{4}=$ $\qquad$
(1995, Chongqing and Four Other Cities Junior High School Mathematics League)
$$
\begin{array}{l}
\because x y=\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{2} \\
\quad=\frac{4-\frac{5}{2}}{2}=\frac{3}{4}, \\
\therefore x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2(x y)^{2} \\
=\left(\frac{5}{2}\right)^{2}-2 \times\left(\frac{3}{4}\right)^{2}=\frac{41}{8} .
\end{array}
$$

The given two equations form a system of equations about $x, y$. Therefore, $x$ and $y$ can be found, but it is more convenient to handle them as a whole without solving for $x$ and $y$ individually.
$$
\begin{array}{l}
\because x y=\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{2} \\
\quad=\frac{4-\frac{5}{2}}{2}=\frac{3}{4}, \\
\therefore x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2(x y)^{2} \\
=\left(\frac{5}{2}\right)^{2}-2 \times\left(\frac{3}{4}\right)^{2}=\frac{41}{8} .
\end{array}
$$