Let the numbers be x and y.

Given that the sum is 20.

==> x + y= 20 ..............(1)

Also, given that the larger number is 4 less than twice the smaller.

==> x = 2y -4 ..............(2)

We will solve the system by substitution.

We will substitute (2) into (1).

==> (2y-4) + y= 20

==> 3y -4 = 20

==> 3y = 24

==> y= 24/3 = 8

==> x = 20-8= 12

**Then, the larger number is 12 and the smaller number is 8.**

Let the smaller number be x, the other number is 4 less than twice the smaller number or 2x - 4

The sum of the two numbers is 20

=> x + 2x - 4 = 20

=> 3x = 24

=> x = 8

The other number is 2*8 - 4 = 12

**The two numbers are 8 and 12.**