Question

the sum of digit of a two digit number is 9 when we add 45 in that number the digit of that number is reversed find the number​

Answer 1

Given:

The sum of 2 digit number=9

Let:

The ones digits be X

The ten's digits be y

The number=10y+X

A.T.Q

Sum of 2 digits=9

==>X+y=9------(1)

When 45 is added to the number that number of digits is reversed

==>10y+X+45=10x+y

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

==>9x-9y=45

Dividing by 9 on both sides

==>X-y=5-------(2)

Now, solving equation I and 2

==>X+y=9

==>x-y=5

Adding equations here

==>2x=14

==>X=7

Putting the value of X=7 in EQ 1

==>X+y=9

==>7+y=9

==>y=9-7

==>y=2

Hence,

The number=10y+X=10×2+7=27

Answer 2

Let the ten's digit and one's digit be x and y respectively.

Given

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

$10x+y+45 = 10y+x\\\implies10x-x+y-10y=-45\\\implies9x-9y=-45\\\implies x-y=-5--(2)$

Solving, 1 and 2, we get,

$x+y=9\\\underline{x-y=-5}\\\underline{\underline{2x=4}}\\\implies x = 2\\\implies y = 7$

$\boxed{\boxed{\bold{Therefore, \ the \ number \ is \ 27}}}}$