Question 1082461
A and B are integers from 0-9. 
Let the original number be {{{10A+B}}}.
The sum of its digits would be {{{A+B}}}.
Digits reversed would be {{{10B+A}}}.
So,
{{{10A+B=7(A+B)}}}
{{{10A+B=7A+7B}}}
{{{3A=6B}}}
1.{{{A=2B}}}
and
{{{10B+A=2(10A+B)-30}}}
{{{10B+A=20A+2B-30}}}
2.{{{19A-8B=30}}}
Substituting from 1 into 2,
{{{19(2B)-8B=30}}}
{{{38B-8B=30}}}
{{{30B=30}}}
Solve for B, then use either equation to solve for A.