Question

Solve each of the following problems by using two variables.

The sum of two digit number and the number obtained by interchanging the digits is 99. If the digits differ by 3, find the number.

Answer 1

Let the unit place of two digit number be $x$ and ten's place be $y$.
$\therefore$ The two digit number is $10y+x$.
If the digits are interchanged then the number becomes, $10x+y$
Now as per condition 1, i.e., The sum of two digit number and the number obtained by interchanging the digits is 99.
The Equation becomes,
$10x+y+10y+x=99$
$\therefore x+y=9$ .....(1)
As per condition 2, i.e., digits differ by 3,
The Equation becomes,
$x-y=3$ .......................(2)
Solving (1) and (2) we get,
$x=6$ and $y=3$
$\therefore$ The number is $36$