The sum of two digits no. …

Suppose, the digit at units and tens place of the given number be x and y respectively.
{tex}\therefore{/tex} the number is {tex}10y + x{/tex}
After interchanging the digits, the number becomes {tex}10x + y{/tex}
Given: The sum of the numbers obtained by interchanging the digits and the original number is 66.
Thus, {tex}(10x + y) + (10y + x) =66{/tex}
{tex}\Rightarrow{/tex} {tex}10x + y + 10y + x = 66{/tex}
{tex}\Rightarrow{/tex} {tex}11x +11y =66{/tex}
{tex}\Rightarrow{/tex} {tex}11(x + y) = 66{/tex}
{tex}\Rightarrow x + y = \frac{{66}}{{11}}{/tex}
{tex}\Rightarrow{/tex} {tex}x + y = 6{/tex} .....(i)
Also given, the two digits of the number are differing by 2.
{tex}\therefore{/tex} we have {tex}x - y = ±2{/tex}....(ii)
So, we have two systems of simultaneous equations,
{tex}x - y = 2, \;x + y = 6{/tex}
{tex}x - y = -2, \;x + y = 6{/tex}
Here x and y are unknowns. We have to solve the above systems of equations for x and y.

1. First, we solve the system
   {tex}x - y = 2{/tex}
   x + y = 6
   Adding the two equations,
   {tex}\Rightarrow(x - y) + (x + y) = 2 + 6{/tex}
   {tex}\Rightarrow{/tex} {tex}x - y + x + y = 8{/tex}
   {tex}\Rightarrow{/tex} {tex}2x = 8{/tex}
   {tex}\Rightarrow x = \frac{8}{2} {/tex}
   {tex}\Rightarrow{/tex}  {tex}x = 4{/tex}
   Substituting the value of x in the first equation, we have
   {tex}4 - y = 2{/tex}
   {tex}\Rightarrow{/tex} {tex}y = 4 - 2{/tex}
   {tex}\Rightarrow{/tex} {tex}y = 2{/tex}
   Hence, the number is 10 {tex}\times{/tex} 2 + 4 = 24
2. Now, we solve the system
   {tex}x - y= -2{/tex}
   {tex}x + y = 6{/tex}
   Adding the two equations, we have
   {tex}(x - y) + (x + y) = -2 + 6{/tex}
   {tex}\Rightarrow{/tex} {tex}x - y + x + y = 4{/tex}
   {tex}\Rightarrow{/tex} {tex}2x = 4{/tex}
   {tex}\Rightarrow x = \frac{4}{2} {/tex}
   {tex}\Rightarrow{/tex} x = 2
   Substituting the value of x in the first equation, 
   {tex}\Rightarrow2 - y = -2{/tex}
   {tex}\Rightarrow{/tex} {tex}y = 2 + 2{/tex}
   {tex}\Rightarrow{/tex} y = 4
   Hence, the number is 10 {tex}\times{/tex} 4 + 2 = 42
   Thus, the two numbers are 24 and 42.