1. Represent the fraction $\frac{44}{55}$ as a sum of two positive fractions whose denominators are 13 and 5.
Solution. From the equation $\frac{44}{55}=\frac{x}{13}+\frac{y}{5}$ we obtain the equation $5 x+13 y=44$, which needs to be solved in the set of natural numbers. From it, we get $x=\frac{44-13 y}{5}=8-2 y+\frac{4-3 y}{5}$. If we set $u=\frac{4-3 y}{5}$, and it is an integer, we get that $y=\frac{4-5 u}{3}=1-2 u+\frac{1+u}{3}$. If we introduce a new substitution $v=\frac{1+u}{3}$, v is an integer, we get that $u=3 v-1$. If we express $x$ and $y$ in terms of $v$, we get $x=13 v+1$ and $y=-5 v+3$. The only solution in the set of natural numbers is obtained for $v=0$, which gives $x=1$ and $y=3$.