Question 485783


let two-digit number be {{{xy}}} where {{{x=tens}}} and {{{y=ones}}}

given:

{{{y=x+3}}}.....1

{{{x+y=13}}}....2

substitute {{{y}}} from 1

{{{x+x+3=13}}}...solve for {{{x}}}

{{{2x=13-3}}}

{{{2x=10}}}

{{{x=5}}}.......tens

{{{y=x+3}}}

{{{y=5+3}}}

{{{y=8}}}.....ones


so, your number is:

{{{58}}}