let us consider the unit place digit as x and tens place digit as y.
The equations becomes 10y + x……..equation (1)
From the question, a two-digit number is 3 more than 4 times the sum of its digits
∴from the above condition, 4(y + x) + 3……… equation (2)
Combining equation 1 and 2
4(y + x) + 3 = 10y + x
4y + 4x + 3 = 10y + x
4x – x + 4y – 10y = -3
3x – 6y = -3
3(x – 2y) = -3
x -2y = -1 …………..equation (3)
From the second condition, If 18 is added to the number, its digits are reversed
∴the reversed number is 10x + y……..equation (4)
∴by the given condition
(10y + x) + 18 = 10x + y
10y – y =10x –x -18
9y – 9x = -18
9(y – x) = -18
y – x = -2 ……….equation (5)
Solving equation 3 and 5 simultaneously we get,
y=3 and x = 5
∴The required number is (10y + x) = (10(3) + 5) = 30 + 5 = 35
