Let's denote the two-digit number as $N$, and let the tens digit be $x$ and the ones digit be $y$.

From the problem statement, we have the following conditions:

1. The digit at the ones place is 4 more than the tens place: $y = x + 4$

2. The ratio of the number $N$ and the number formed by reversing its digits is 47. The number $N$ can be written as: $N = 10x + y$ The number formed by reversing the digits would be: $\text{Reversed number} = 10y + x$ The given ratio is: $\frac{N}{\text{Reversed number}} = 47$ Substituting the expressions for $N$ and the reversed number: $\frac{10x + y}{10y + x} = 47$

Now, substitute $y = x + 4$ into this equation: $\frac{10x + (x + 4)}{10(x + 4) + x} = 47$ Simplifying the numerator and denominator: $\frac{10x + x + 4}{10x + 40 + x} = 47$ $\frac{11x + 4}{11x + 40} = 47$

Now, cross-multiply to solve for $x$: $11x + 4 = 47(11x + 40)$ $11x + 4 = 517x + 1880$ $4 - 1880 = 517x - 11x$ $-1876 = 506x$ $x = \frac{-1876}{506} = -3.71$

It seems there was a miscalculation in some part of the problem. Let me verify this!