A two-digit

Let tens place digit be y and the units place be x.

∴ Our number is (10y + x)

Our given first condition is that our number is 3 more than 4 times the sum of its digits.

∴ By given condition,

4(y + x) + 3 = (10 y + x)
4y + 4x + 3 = 10y + x
6y - 3x = 3
3(2y - x) = 3
2y - x =1 …(1)

Our given second condition is that if 18 is added to the number, its digits are reversed.

The reversed number is (10x + y)

∴ By given condition,

(10y + x) + 18 = 10x + y
10y - y + x -10x = -18
9y - 9x = -18
9(y - x) = -18
y - x = -2
y = x - 2     [2]

Putting this value of 'y' in eq (1), we have
2(x - 2) - x = 1
2x - 4 - x = 1
x = 5

From [2], we have
y = 5 - 2 = 3

Hence, 

y = 3 and x = 5

∴ Our number = (10 × 3 + 5) = 35

Hence, our number is 35.