755. "Forty mice were walking, carrying forty groats,
Two poorer mice carried two groats each, Many mice - with no groats at all. The bigger ones carried seven each. And the rest carried four each.

How many mice walked without groats?"
$\triangle$ Let the number of mice that walked without groats be denoted by $x$, the number of large mice by $y$, and the number of those that carried four groats by $z$. We can set up a system of equations with the unknowns $x, y$, and $z$:

$$
\left\{\begin{array} { l } 
{ 2 \cdot 2 + 7 y + 4 z = 4 0 , } \\
{ 2 + x + y + z = 4 0 , }
\end{array} \quad \left\{\begin{array}{l}
7 y+4 z=36 \\
x+y+z=38
\end{array}\right.\right.
$$

This is a system of two linear equations with three unknowns.

To solve this system, note that from the first equation, $y$ must be divisible by 4. Additionally,

$$
7 y<7 y+4 z, \quad 7 y<36
$$

from which $y \leqslant 5$. There is only one such $y$: $y=4$. Then, from the first equation, we find $z$, and subsequently from the second equation, we find $x$:

$$
7 \cdot 4+4 z=36, \quad 4 z=8, \quad z=2 ; \quad x+4+2=38, \quad x=32
$$

Answer: 32.