A two digit no. Is such …

Suppose that the digits at units and tens place of the given number be x and y respectively.
Thus, the number is {tex}10y + x.{/tex}
The product of the two digits of the number is 20.
Thus, we have {tex}xy = 20{/tex}
After interchanging the digits, the number becomes {tex}10x + y{/tex}
If 9 is added to the number, the digits interchange their places.
Thus, we have
{tex}(10y + x) + 9 = 10x + y{/tex}
{tex}\Rightarrow{/tex} {tex}10y + x + 9 = 10x+ y{/tex}
{tex}\Rightarrow{/tex} {tex}10x + y - 10y - x = 9{/tex}
{tex}9x - 9y = 9{/tex}
{tex}\Rightarrow{/tex}{tex} 9(x - y) = 9{/tex}
{tex}\Rightarrow x - y = \frac{9}{9}{/tex}
{tex}\Rightarrow{/tex} {tex}x - y = 1{/tex}
So, we have the systems of equations
{tex}xy = 20{/tex} ....(i)
{tex}x - y = 1{/tex} ....(ii)
Here x and y are unknowns.
We have to solve the above systems of equations for x and y.
Substituting {tex}x = 1 + y{/tex} from the second equation to the first equation, we get {tex}(1+ y) y = 20{/tex}
{tex}\Rightarrow{/tex} {tex}y + y^2 = 20{/tex}
{tex}\Rightarrow{/tex} {tex}y^2 + y - 20 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}y^2 + 5y - 4y - 20 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}y(y + 5) - 4(y + 5) = 0{/tex}
{tex}\Rightarrow{/tex} {tex}(y + 5)(y - 4) = 0{/tex}
{tex}\Rightarrow{/tex} {tex}y = -5\ or\ y = 4{/tex}
Substituting the value of y in the second equation, we have

Note that in the first pair of solution the values of x and y are both negative. But the digits of the number can't be negative. So, we must remove this pair.
Hence, the number is 10 {tex}\times{/tex} 4 + 5 = 45