Question

the sum of the digits of a two digit number is 9 also nine times this number is twice the number obtained by reversing the order of the digits find the number

Answer 1

let  two digit number be 10x + y

sum of two digit = x+y = 9 

reversed number = 10y+x

according to q.

9(10x + y) = 2(10y + x)

90x + 9y = 20x + 2y

88x  =  11y

8x = y

putting value in x  + y = 9

x+ 8x =9

9x = 9

 x = 1 

y  = 9 - x

y  = 8

number is 10x + y

10 + 8

18 

Answer 2

Here is your solution

Given :-

The sum of the digits of a two digit number is 9

Let
The digits be x and y, and the number be 10x + y.

I. e

x + y = 9 ------ (1)

Now
A/q

9×(10x + y) = 2×(10y + x)

90x + 9y = 20y + 2x

88x = 11y

y = 8x ---- (2)

putting the value of y in equation (1.) we get,

x + 8x = 9
9x = 9.
x = 1

y = 8×1 = 8.

Hence

The number is 10x +y = 10×1 + 8 = 18.

Hope it helps you