Let the two-digit number be represented as \( 10t + u \), where \( t \) is the tens digit and \( u \) is the units digit.

We are given two conditions:

1. Three times the tens digit is 2 less than twice the units digit.
\[
3t = 2u - 2 \tag{1}
\]

2. Twice the number is 20 greater than the number obtained by reversing the digits.
\[
2(10t + u) = 20 + (10u + t) \tag{2}
\]

Step 1: Simplify Equation (2)
Expanding both sides:
\[
20t + 2u = 20 + 10u + t
\]
Now, rearrange the terms:
\[
20t - t + 2u - 10u = 20
\]
\[
19t - 8u = 20 \tag{3}
\]

Step 2: Solve the system of equations
From Equation (1):
\[
3t = 2u - 2 \implies 2u = 3t + 2 \implies u = \frac{3t + 2}{2} \tag{4}
\]
Substitute this into Equation (3):
\[
19t - 8\left( \frac{3t + 2}{2} \right) = 20
\]
Simplify:
\[
19t - 4(3t + 2) = 20
\]
\[
19t - 12t - 8 = 20
\]
\[
7t = 28 \implies t = 4
\]

Step 3: Find the value of \( u \)
Substitute \( t = 4 \) into Equation (4):
\[
u = \frac{3(4) + 2}{2} = \frac{12 + 2}{2} = \frac{14}{2} = 7
\]

Step 4: Form the number
The number is \( 10t + u = 10(4) + 7 = 47 \).

Thus, the two-digit number is \( \boxed{47} \).