Question 929228
Let {{{x}}} = one's digit in original number
Let {{{y}}} = ten's digit in original number
value of original number = {{{10y + x}}}

In new number (with reversed digits):
{{{y}}} = one's digit
{{{x}}} = ten's digit
value of new number = {{{10x + y}}}

given: The sum of the digits of a certain two-digit number is {{{12}}}. Reversing its digits decreases the number by {{{18}}}.

We can now form these equations:
"Sum of digits of two-digit no. is {{{12}}}" (original number)

{{{y + x = 12}}}...eq.1

"Reverse the digits, the no. is decreased by {{{18}}}."

{{{10y + x - 18 = 10x + y}}}

{{{10y - y + x -18 = 10x + y - y}}}

{{{9y + x -18 = 10x}}}

{{{9y + x - x -18 = 10x - x}}}

{{{9y -18= 9x}}}

{{{9y/9 -18/9 = 9x/9}}}

{{{y -2= x}}}....eq.2

Now compare the two equations:

{{{y + x = 12}}}
{{{y -2= x}}}

We can substitute {{{(y -2)}}} for {{{x}}} in the first equation to solve:

{{{y + (y -2) = 12}}}

{{{2y -2= 12}}}

{{{2y + 2 - 2 = 12 +2}}}

{{{2y = 14}}}

{{{2y/2 = 14/2}}}

{{{highlight(y =7)}}}

Now substitute this value for {{{y}}} into one of the equations to find {{{x}}}:

{{{y -2 = x}}}

{{{7-2 = x}}}

{{{highlight(x = 5)}}}

Our original number is  {{{10y + x=highlight(75)}}}.