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

Writing $a=21r$, the second condition becomes $r\equiv 4(\mathrm{mod}5)$. Writing $b=15s$, its second condition becomes $s\equiv 6(\mathrm{mod}7)$. Consequentlygives an infinite set of positive integer solutions. Conversely these are the positive solutions in the indicated congruence classes, by the Chinese remainder theorem.