Topic: Simultaneous equations Class 9 ICSE​

Denote the number by 10a + b, where both a and b are integers picked from {1, 2, 3, …, 9}.

"the difference of whose digits is 3"   ==>   |a - b| = 3

"4 times the number is equal to 7 times the number obtained by reversing the digits"   ==>   4 (10a + b) = 7 (10b + a)

Simplifying the second equation, you get

40a + 4b = 70b + 7a

33a - 66b = 0

a - 2b = 0

Assume a > b, in which case we would have |a - b| = a - b, so

a - b = 3

Eliminate a to solve for b, then for a :

(a - b) - (a - 2b) = 3 - 0

b = 3   ==>   a = 6

Then the original number is 63.

If instead we had assumed a < b, we would have had |a - b| = b - a = 3. Then

2 (b - a) + (a - 2b) = 2×3 + 0

-a = 6

But neither a nor b can be negative, so this case is moot.