A two-digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.

#### Solution

Let the digits at units and tens place of the given number be *x* and *y* respectively. Thus, the number is ` 10y + x`.

The number is 3 more than 4 times the sum of the two digits. Thus, we have

` 10 y + x = 4(x +y)+ 3`

` ⇒ 10 y + x = 4x + 4y + 3`

` ⇒ 4x + 4y -10y -x =-3`

` ⇒ 3x - 6y = -3`

` ⇒ 3 ( x - 2 y)= -3`

` ⇒ x - 2y = -3/3`

` ⇒ x - 2y = -1`

After interchanging the digits, the number becomes `10 x + y.`.

If 18 is added to the number, the digits are reversed. Thus, we have

` ( 10 y + x )+ 18 = 10x + y`

` ⇒ 10x + y -10y -x =18`

` ⇒ 9x -9y = 18`

` ⇒ 9( x - y)=18`

` ⇒ x -y = 18 /9`

` ⇒ x - y =2`

So, we have the systems of equations

` x - 2y =-1`

` x - y =2`

Here *x* and *y* are unknowns. We have to solve the above systems of equations for *x*and *y*.

Subtracting the first equation from the second, we have

` ( x - y)-(x - 2y )=2 -(-1)`

` ⇒ x - y -x + 2y =3`

` ⇒ y = 3`

Substituting the value of *y *in the first equation, we have

` x - 2xx3 =-1`

`⇒ x - 6 = -1 `

` ⇒ x = -1+6`

` ⇒ x = 5`

Hence, the number is ` 10 xx3 + 5 = 35`