Writing $a=21r$, the second condition becomes $r\equiv4\pmod5$. Writing $b=15s$, its second condition becomes $s\equiv6\pmod7$. Consequently
$$
\boxed{a=84+105m,\qquad b=90+105n,\qquad m,n\in\mathbb Z_{\geq0}}
$$
gives an infinite set of positive <integer> solutions. Conversely these are the positive solutions in the indicated <congruence classes>, by the <Chinese remainder theorem>.

Solution

Past exam of the short options course of the University of Oxford / 2014 / 1 / b / i / Solution

 Solution (source code)