, 2023

Since $a$ and $b$ are positive integers, then each of
$\dfrac{a}{7}$ and $\dfrac{2}{b}$ is greater than 0.

The sum of $\dfrac{a}{7}$ and $\dfrac{2}{b}$ is equal to 1, and so each
is less than 1.

Since $\dfrac{a}{7}$ is greater than
0 and less than 1, then the possible values of $a$ are $1,2,3,4,5,6$.

By substituting each of these values for $a$ into the equation one at a time, we can
determine if there is a positive integer value of $b$ for which the equation is true.

Substituting $a=1$, we get $\dfrac{1}{7}+\dfrac{2}{b}=1$ or $\dfrac{2}{b}=1-\dfrac{1}{7}$, and so $\dfrac{2}{b}=\dfrac{6}{7}$.

Since $\dfrac{2}{b}=\dfrac{6}{7}$,
we can multiply the numerator and denominator of the first fraction by 3
(which is $6\div2$) to get $\dfrac{6}{3b}=\dfrac{6}{7}$.

This gives $3b=7$ which does not
have an integer solution ($b=\dfrac73$).

Thus when $a=1$, there is no positive
integer value of $b$ that satisfies
the equation.

Substituting $a=2$, we get $\dfrac{2}{7}+\dfrac{2}{b}=1$ or $\dfrac{2}{b}=1-\dfrac{2}{7}$, and so $\dfrac{2}{b}=\dfrac{5}{7}$.

Since $\dfrac{2}{b}=\dfrac{5}{7}$, we
can multiply the numerator and denominator of the first fraction by 5,
and the numerator and denominator of the second fraction by 2 to get
$\dfrac{10}{5b}=\dfrac{10}{14}$.

This gives $5b=14$ which does not
have an integer solution.

Thus when $a=2$, there is no positive
integer value of $b$ that satisfies
the equation.

Substituting $a=3$, we get $\dfrac{3}{7}+\dfrac{2}{b}=1$ or $\dfrac{2}{b}=1-\dfrac{3}{7}$, and so $\dfrac{2}{b}=\dfrac{4}{7}$.

Since $\dfrac{2}{b}=\dfrac{4}{7}$, we
can multiply the numerator and denominator of the first fraction by 2 to
get $\dfrac{4}{2b}=\dfrac{4}{7}$.

This gives $2b=7$ which does not have
an integer solution. Thus when $a=3$,
there is no positive integer value of $b$ that satisfies the equation.

Substituting $a=4$ and
simplifying, we get $\dfrac{2}{b}=\dfrac{3}{7}$.

Since $\dfrac{2}{b}=\dfrac{3}{7}$, we
can multiply the numerator and denominator of the first fraction by 3,
and the numerator and denominator of the second fraction by 2 to get
$\dfrac{6}{3b}=\dfrac{6}{14}$.

This gives $3b=14$ which does not
have an integer solution.

Thus when $a=4$, there is no
positive integer value of $b$ that
satisfies the equation.

Substituting $a=5$ and
simplifying, we get $\dfrac{2}{b}=\dfrac{2}{7}$.

Since the numerators are equal, then the denominators must be equal, and
so $b=7$ satisfies the equation.

Finally, substituting $a=6$ and
simplifying, we get $\dfrac{2}{b}=\dfrac{1}{7}$.

Since $\dfrac{2}{b}=\dfrac{1}{7}$, we
can multiply the numerator and denominator of the second fraction by 2
to get $\dfrac{2}{b}=\dfrac{2}{14}$,
and so $b=14$.

Thus, there are two pairs of positive integers $a$ and $b$ that satisfy the given equation: $a=5, b=7$ and $a=6, b=14$.