Question 957727
Let {{{ a }}} = the tens digit
Let {{{ b }}} = the units digit
------------------------
(1) {{{ 5*( a + b ) = 10a + b - 13 }}}
(2) {{{ 4*( b + a ) = 10b + a - 21 }}}
--------------------------------
(1) {{{ 5a + 5b = 10a + b - 13 }}}
(1) {{{ 5a - 4b = 13 }}}
---------------------
(2) {{{ 4b + 4a  = 10b + a - 21 }}}
(2) {{{ -3a + 6b = 21 }}}
----------------------
Multiply both sides of (2) by {{{ 5 }}}
and both sides of (1) by {{{ 3 }}}
and add the equations
(1) {{{ 15a - 12b = 39 }}}
(2) {{{ -15a + 30b = 105 }}}
-------------------------
{{{ 18b = 144 }}}
{{{ b = 8 }}}
and
(1) {{{ 5a - 4*8 = 13 }}}
(1) {{{ 5a = 13 + 32 }}}
(1) {{{ 5a = 45 }}}
(1) {{{ a = 9 }}}
The original number is 98
Number with reversed digits is 89
{{{ 98 - 89 = 9 }}}
-----------------
check:
(1) {{{ 5*( a + b ) = 10a + b - 13 }}}
(1) {{{ 5*( 9 + 8 ) = 10*9 + 8 - 13 }}}
(1) {{{ 5*17 = 98 - 13 }}}
(1) {{{ 85 = 85 }}}
OK
(2) {{{ 4*( b + a ) = 10b + a - 21 }}}
(2) {{{ 4*( 8 + 9 ) = 10*8 + 9 - 21 }}}
(2) {{{ 4*17 = 89 - 21 }}}
(2) {{{ 68 = 68 }}}
OK