By cross multiplication method, find such a two digit number such that, the digit at unit's place is twice the digit at tens place and the number obtained by interchanging the digits of the number is 36 more than the original number.

Let the required two digit number be (10x + y) where x is the digit at tens place and y is the digit at unit place.

On interchanging the digits the two digit number formed will be (10y + x).

According to the question,

y = 2x and (10y + x) = 36 + (10x + y)

⇒ 2x – y = 0 and 10x + y – 10 y – x + 36 = 0

⇒ 2x –y = 0 and 9x – 9y + 36 = 0

⇒ 2x – y = 0 and x – y + 4 = 0

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0we get,

a_{1 =} 2, b₁ = – 1, c₁ = 0; a₂ = 1, b₂ = – 1, c₂ = 4

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (4, 8).

So, the required two digit number will be (10x + y) = (10× 4 + 8) = 48