The number of ordered pairs $(x,y)$ satisfying the equation The number of ordered pairs $(x,y)$ satisfying the equation $\lfloor\frac{x}{2}\rfloor+\lfloor\frac{2x}{3}\rfloor+\lfloor\frac{y}{4}\rfloor+\lfloor\frac{4y}{5}\rfloor=\frac{7x}{6}+\frac{21y}{20}$,where $0<x,y<30$

It appears that $\frac{x}{2}+\frac{2x}{3}=\frac{7x}{6}$ and $\frac{y}{4}+\frac{4y}{5}=\frac{21y}{20}$ but $\frac{x}{2}$ and $\frac{2x}{3}$ being inside the floor function,i can not add them up directly,same problem here,i cannot add $\frac{y}{4}$ and $\frac{4y}{5}$ because they are inside the floor function.What should i do to solve it?

Note that
$$\Bigl\lfloor\frac x2\Bigr\rfloor\le\frac x2\ ,$$
and likewise for the other terms.  So we always have $LHS\le RHS$, and the only way they can be equal is if
$$\Bigl\lfloor\frac x2\Bigr\rfloor=\frac x2\ ,$$
and likewise for the other terms.  So
$$\frac x2\ ,\quad \frac{2x}3\ ,\quad \frac y4\ ,\quad\frac{4y}5$$
must all be integers.  Can you take it from here?
