Question

the sum of the digit of the eight two number is 11. When the number are reversed the number is 9 find the original number

Answer 1

Let the digit in the number is xy

So number is 10x + y

Given, sum of the digits of a two-digit number is 9

=> x + y = 11 ..........................1

Again

     10y + x = 10x + y - 45

=> 10x + y - 10y - x = 45

=> 9x - 9y = 45

=> 9(x - y) = 45

=> x - y = 45/9

=> x - y = 5  ..................2

Add equation 1 and 2, we get

      x + y + x - y = 11 + 5

=> 2x = 16

=> x = 16/2

=> x = 8

Now, from equation 1, we get

=> 8 + y = 11

=> y = 11 - 8

=> y = 3

Answer 2

From x + y = 15, we solve for one of them, say   x = 15-y,   then substitute that into the other equation:
      10y + (15-y) = 9 + 10(15-y) + y                    (now it’s all y, so we can solve for y)
        9y + 15   =   159 - 9y
             18y = 144
               y = 8

The original number was 78.

 Now, plug that value back into either equation (x+y=15 is the easiest):
             x + 8 = 15
             x = 7