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.
$$\begin{array}{}\text{(1)}& 3t=2u-2\end{array}$$

2. Twice the number is 20 greater than the number obtained by reversing the digits.
$$\begin{array}{}\text{(2)}& 2(10t+u)=20+(10u+t)\end{array}$$

Step 1: Simplify Equation (2)
Expanding both sides:
$$20t+2u=20+10u+t$$
Now, rearrange the terms:
$$20t-t+2u-10u=20$$
$$\begin{array}{}\text{(3)}& 19t-8u=20\end{array}$$

Step 2: Solve the system of equations
From Equation (1):
$$\begin{array}{}\text{(4)}& 3t=2u-2⟹2u=3t+2⟹u=\frac{3t+2}{2}\end{array}$$
Substitute this into Equation (3):
$$19t-8\left(\frac{3t+2}{2}\right)=20$$
Simplify:
$$19t-4(3t+2)=20$$
$$19t-12t-8=20$$
$$7t=28⟹t=4$$

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 $\boxed{{\displaystyle 47}}$.