4. Find the area of the region enclosed by the graph of the equation $|x-60|+|y|=\left|\frac{x}{4}\right|$.

 untranslated text remains the same as requested.
4. First, the graph of this equation is symmetric about the $x$-axis, so we only need to find the area enclosed by
$$
\left\{\begin{array}{l}
\left.y=\left|\frac{x}{4}\right|-1 x-60 \right\rvert\,, \\
y \geqslant 0
\end{array}\right.
$$

The region enclosed by the difference of $0 \cdots \times 1$ and the $x$-axis in the upper half-plane is the triangle $\triangle D A B$. It is easy to find that the coordinates of the three vertices are $D$ (60, 15), $A(48,12), B(80,20)$. Thus, $S_{\triangle D A B}=\frac{A B \times C D}{2}=240$. Therefore, the area of the region enclosed by the graph of the original equation is 480.