[Algebra]
Problem: Determine all pairs $(x, y)$ of real numbers that satisfy the system of equations

$$
\begin{aligned}
x\left(\frac{1}{2}+y-2 x^{2}\right) & =0 \\
y\left(\frac{5}{2}+x-y\right) & =0
\end{aligned}
$$

Solution: From the first equation, $x\left(\frac{1}{2}+y-2 x^{2}\right)=0$, we obtain $x=0$ or $\frac{1}{2}+y-2 x^{2}=0$.

From the second equation, $y\left(\frac{5}{2}+x-y\right)=0$, we obtain $y=0$ or $\frac{5}{2}+x-y=0$.

If $x=0$, the first equation is satisified.

For the second equation to be true in this case, we need $y=0$ (giving the solution $(0,0)$ ) or $\frac{5}{2}+0-y=0$. The second equation gives $y=\frac{5}{2}$ (giving the solution $\left(0, \frac{5}{2}\right)$ ).

If $y=0$, the second equation is satisified.

For the first equation to be true in this case, we need $x=0$ (giving the solution $(0,0)$ ) or $\frac{1}{2}+0-2 x^{2}=0$. The second equation gives $x^{2}=\frac{1}{4}$ or $x= \pm \frac{1}{2}$ (giving the solutions $\left(\frac{1}{2}, 0\right)$ and $\left.\left(-\frac{1}{2}, 0\right)\right)$.

So far, we have accounted for all solutions with $x=0$ or $y=0$.

If $x \neq 0$ and $y \neq 0$, then for both equations to be true, we need $\frac{1}{2}+y-2 x^{2}=0$ (or $1+2 y-4 x^{2}=0$ ) and $\frac{5}{2}+x-y=0$ ( or $5+2 x-2 y=0$ ).

Adding these two equations, we obtain $6+2 x-4 x^{2}=0$.

This is equivalent to $2 x^{2}-x-3=0$ or $(2 x-3)(x+1)=0$, whose solutions are $x=\frac{3}{2}$ and $x=-1$.

The equation $\frac{5}{2}+x-y=0$ tells us that $y=x+\frac{5}{2}$.

If $x=\frac{3}{2}$, then $y=4$; if $x=-1$, then $y=\frac{3}{2}$.

Therefore, the complete list of pairs that satisfy the given system of equations is

$$
(x, y)=(0,0),\left(0, \frac{5}{2}\right),\left(\frac{1}{2}, 0\right),\left(-\frac{1}{2}, 0\right),\left(\frac{3}{2}, 4\right),\left(-1, \frac{3}{2}\right)
$$

Final answer: $(0,0),(0, \frac{5}{2}),(\frac{1}{2}, 0),(-\frac{1}{2}, 0),(\frac{3}{2}, 4),(-1, \frac{3}{2})$