Question

sum of the digits of a two 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 unit digit  = x

Tens digit  = y

Number will 10 times the tens digit + unit times the unit digit

Hence number will 10 y + x


Sum of digits are 9

So that

X + y  = 9        ………….(1)

 nine times this number is twice the number obtained by reversing the order of the digits


9 (10 y + x )  =  2 (10 x + y )

90 y + 9 x  = 20 x + 2y

88 y – 11 x  = 0

Divide by 11 we get

8 y  - x  = 0     …………..(2)

X + y  = 9        ………….(1)


Adding both equations we get

9 y = 9

Y = 9/9 = 1

Plug this value in equation first we get

X+ y = 9

X + 1 = 9

X = 8

So our original number is 10 y  + x    = 10*1 + 8 = 18



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Answer 2

Step-by-step explanation:

let the number be 10x +y

then,x + y = 9 .....

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

= 90x + 9y = 20y +2x

= 90x -2x +9y -20y = 0

= 88x -11y = 0

= 8x -y = 0.............(2)

adding (1) and (2),

9x = 9

= x = 1

= x = 1

putting this in eqn (2),

8×1- y = 0

= y = 8

hence, no. = 10×1+8= 18