Question 185238
Let {{{a}}}= the larger number 
Let {{{b}}}= the smaller number
given:
(1) {{{b = a - 7}}}
(2) {{{4b + 2a = 62}}}
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You have 2 equations and 2 unknowns. In all
cases where the number of unknowns = the number of equations,
the problem is solvable.
I'll substitute (1) in (2)
(2) {{{4*(a - 7) + 2a = 62}}}
(2) {{{4a - 28 + 2a = 62}}}
(2) {{{6a = 90}}}
(2) {{{a = 15}}}
Now plug this result back into (1)
(1) {{{b = 15 - 7}}}
(1) {{{b = 8}}}
The larger number is 15 and the smaller is 8
check answer:
(2) {{{4b + 2a = 62}}}
(2) {{{4*8 + 2*15 = 62}}}
(2) {{{32 + 30 = 62}}}
(2) {{{62 = 62}}} 
OK