Question 161736
Let's call the number in the tens place A and the number in the ones place B.
The number would be AB or 10*A+B.
1.{{{10A+B=4*(A+B)}}}
If the number is reversed then,
2.{{{10B+A=27+(10A+B)}}}
Let's simplify both equations,
1.{{{10A+B=4A+4B}}}
1.{{{6A-3B=0}}}
and now eq. 2,
2.{{{10B+A=27+(10A+B)}}}
2.{{{-9A+9B=27}}}
Let's multiply eq. 1 by 3 and add to eq. 2,
then we can solve for A.
{{{18A-9B+(-9A)+9B=0+27}}}
{{{9A=27}}}
{{{A=3}}}
Now use eq. 1 to solve for B.
1.{{{6A-3B=0}}}
{{{18-3B=0}}}
{{{-3B=-18}}}
{{{B=6}}}
The original number is 36, the reversed order number is 63.
Check the answer.
36+27=63
63=63.
True statement.
Good answer.