jmc

To work with the absolute values, we take cases on the value of $x$:

For $x<0,$ we have $(60-x)+|y|=-\frac{x}{4},$ or $|y|=\frac{3x}{4}-60.$ But $|y|$ is always nonnegative, whereas $\frac{3x}{4}-60<-60$ whenever $x<0.$ So no part of the graph of the given equation has $x<0.$

For $0\le x<60,$ we have $(60-x)+|y|=\frac{x}{4},$ or $|y|=\frac{5x}{4}-60.$ Since $\frac{5x}{4}-60\ge 0$ when $x\ge 48,$ the graph of the equation consists of two line segments, one from $(48,0)$ to $(60,15),$ and another from $(48,0)$ to $(60,-15).$

For $60\le x,$ we have $(x-60)+|y|=\frac{x}{4},$ or $|y|=-\frac{3x}{4}+60.$ Since $-\frac{3x}{4}+60\ge 0$ when $x\le 80,$ the graph of this equation consists of two line segments, one from $(60,15)$ to $(80,0),$ and another from $(60,-15)$ to $(80,0).$

Altogether, the graph of this equation is a kite, with diagonals of length $80-48=32$ and $15-(-15)=30.$ Therefore, the area of the enclosed region is $\frac{1}{2}\cdot 32\cdot 30=\boxed{480}.$