Question 226203
Let the units digit  be{{{u}}}
Let the tens digit be {{{t}}}
given:
(1) {{{10t + u = 9*(t + u)}}}
(2) {{{10u + t = 10t + u - 63}}}
-------------------------
(1) {{{10t + u = 9t + 9u)}}}
{{{t = 8u}}}
and
(2) {{{10u + t = 10t + u - 63}}}
{{{10u + 8u = 10*8u + u - 63}}}
{{{18u = 81u - 63}}}
{{{81u - 18u = 63}}}
{{{63u = 63}}}
{{{u = 1}}}
and
{{{t = 8u}}}
{{{t = 8*1}}}
{{{t = 8}}}
The original number is 81
check:
(1) {{{10t + u = 9*(t + u)}}}
{{{10*8 + 1 = 9*(8 + 1)}}}
{{{81 = 9*9}}}
{{{81 = 81}}}
(2) {{{10u + t = 10t + u - 63}}}
{{{10*1 + 8 = 10*8 + 1 - 63}}}
{{{18 = 81 - 63}}}
{{{18 = 18}}}
OK