Question

The sum of a two digit number and the number obtained by reversing the order of digits is 99. If the digits of the number differ by 3. find the number

Answer 1

let the digit lf 2 digit no. be x and y
where x in tenth place and y in once
ATQ
CASE:1
(10x+y)+(10y+x) = 99
11y+ 11 x = 99
11 ( x + y) = 99
x+y = 9----------(1)
CASE :2
X-Y = 3----------(2)

by adding them
2x = 12
x = 6

put x = 6
6 - y = 3
y =3

Answer 2

Here is your answer by Sujeet
let to be the 10'unit digit x
""""""""""""""" 1'unit digit y
then,

required no be 10x+y
reverse no be 10y+x
A/q
10x+y+10y+x=99
11x+11y=99
take common,
11(x+y)=99
x+y=9. -------(1)


again,
x-y=3

solve the question by elimination method,
x+y=9
x-y=3
----------
2y=6
y=6/2
y=3

then.
x-y=3
x-3=3
x=3+3
x=6

Again required no be 10x+y
10*6+3
60+3
63
that's all
mark brainliest