Question

The sum of a two-digit number and the number formed by interchanging its digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of the digits in the first number. Find the first  number.

Answer 1

10x+y+10y+x=110
> 11x+11y=110
> x+y=10 ----------1
10x+y-10=4+5x+5y
>5x-4y=14 --------2
(1) * 4
=4x+4y=40 --------1*
1*+2
=4x+4y=40
+5x-4y=14
=9x=54
 therefore, x=6
substituting for x in 1
y=10-6 = 4
therefore: x=6 and y=4 so the first number is 64.

Answer 2

Here is your solution

Let the digit at ones place = y
Let the digit at tens place = x

original number = 10x + y 

Number formed by reversing the digits =10y + x

On adding original number and Reversed number.

10x+y+10y+x = 110
11x + 11y = 110
11 (x + y ) = 110
x + y = 10
x = 10 - y ................................................. (1)

on subtracting 10 from the original number , we get = 10x + y -10
and

4 more than 5 times the sum of digits = 5(x + y) + 4

A/q
 
10x + y -10 = 5(x + y) + 4
10x + y -10 = 5x +5y + 4
10x -5x + y -5y = 10 +4
5x - 4y = 14.................................................. (2)

putting the value of x in equation (2)

5x - 4y = 14
5×(10-y) - 4y = 14
50-5y-4y =14
-9y = 14 -50
-9y = -36
y = 4

Now

x = 10-y
x = 10 - 4
x = 6

original number or first number = 10x + y  = 10×6+4 = 64

Hope it helps you