Question 197725
Let {{{a}}}= the ones digit
Let {{{b}}}= the tens digit
{{{a = b + 1}}}
The number is {{{10b + a}}}
{{{10b + a + 2*(10a + b) = 153}}}
{{{10b + a + 20a + 2b = 153}}}
{{{21a + 12b = 153}}}
Now substitute
{{{21*(b + 1) + 12b = 153}}}
{{{21b + 21 + 12b = 153}}}
{{{33b = 132}}}
{{{b = 4}}}
and
{{{a = b + 1}}}
{{{a = 4 + 1}}}
{{{a = 5}}}
The number is 45
check:
{{{10b + a + 2*(10a + b) = 153}}}
{{{45 + 2*54 = 153}}}
{{{45 + 108 = 153}}}
{{{153 = 153}}}
OK