Task 5. The fraction $\frac{59}{143}$ should be represented as the sum of two proper irreducible fractions.
Solution. The number 143 can be written as $143=11 \cdot 13$. According to this, the fraction $\frac{59}{143}$ can be represented in the form $\frac{59}{143}=\frac{59}{11 \cdot 13}=\frac{x}{11}+\frac{y}{13}$. If we reduce the right side of the last equality to the least common denominator, we get $\frac{13 x+11 y}{143}=\frac{59}{143}$. Two fractions that have the same denominator are equal if and only if they have the same numerator. Since the denominators of the two fractions in the last equality are the same, they will be equal if their numerators are equal. If we equate the numerators, we get the equation $13 x+11 y=59$.

By direct search, it is found that the only solution to the last equation is $x=2, y=3$. Therefore, $\frac{59}{143}=\frac{2}{11}+\frac{3}{13}$.

## VI Section