Find a two-digit number such that three times the tens digit is 2 less than the unit's digit and twice the number is 20 greater than the number obtained by reversing the digits.

Let's name these two digits as a (tens digit) and b (units digit) and then rephrase the problem statement using a and b.

The part of the sentence "three times the tens digit is 2 less than twice the units digits" translates into:

3a = 2b - 2

Considering that both a and b have to be integer numbers less than 10 and that a will have to be an even number

(since 2b - 2 is always even and for 3a to be even, a will have to be even),

the only possible options that will satisfy this equation are: 47 and 24.

The second part of the problem statement does not make much sense.

What exactly is twice the number obtained by reversing the digits?

In case of 47, twice the number obtained by reversing the digits is 2*74 = 148.

In case of 24, twice the number obtained by reversing the digits is 2*42 = 84.