on reversing the digits of a two digit number, number obtaned is 9 less than three times the original number. If difference of these two numbers is 45, find the original number

let the ones place digit be y & tens place digit be x.
the original number = 10x+y
reversed number = 10y+x

according to question,
a) 3(10x+y)-9 = 10y+x
=> 30x+3y-9-10y-x= 0
=> 29x-7y-9 = 0-----(1)

b) as the reversed number is bigger, so
10y+x-(10x+y)=45
=>10y-y-10x+x-45 = 0
=> 9y-9x-45=0
=> y-x-5=0
=> y= x+5 -----(2)

now put equation (2) in equation (1)

29x-7(x+5)-9=0
=> 29x-7x-35-9=0
=> 22x=44
=> x = 2

now put the x=2 in equation (2)
y= 2+5= 7

therefore the original number is 27 & the reversed number is 72.
Thanks...