Are real numbers.

To determine if R is reflexive, we need to check if (x,x) is in R for every x in the domain.

Let's substitute x = x into the equation: 2x - 3x = 20.
Simplifying, we get -x = 20, which is not true for any real number x.

Since there is no real number x that satisfies the equation, R is not reflexive.

To determine if R is symmetric, we need to check if whenever (x,y) is in R, then (y,x) is also in R.

Let's assume (x,y) is in R, which means 2x - 3y = 20.
Rearranging the equation, we get 2x = 3y + 20.
Dividing both sides by 2, we have x = (3y + 20)/2.

Now let's substitute y = x into the equation: 2y - 3x = 20.
Substituting x with (3y + 20)/2, we get 2y - 3((3y + 20)/2) = 20.
Simplifying, we have 2y - (9y + 60)/2 = 20.
Multiplying through by 2, we get 4y - 9y - 60 = 40.
Combining like terms, we have -5y = 100.
Dividing by -5, we get y = -20.

So we have found a solution to the equation 2y - 3x = 20, which is (x,y) = ((3(-20) + 20)/2, -20) = (-10, -20).
Therefore, R is symmetric because whenever (x,y) is in R, (y,x) is also in R.

To determine if R is transitive, we need to check if whenever (x,y) is in R and (y,z) is in R, then (x,z) is in R.

Assume (x,y) is in R and (y,z) is in R, which means 2x - 3y = 20 and 2y - 3z = 20.
We can add these two equations together to get 2x - 3y + 2y - 3z = 40.
Simplifying, we have 2x - y - 3z = 40.

Now, let's substitute y = (2x - 20)/3 into the equation: 2x - ((2x - 20)/3) - 3z = 40.
Multiplying through by 3, we get 6x - 2x + 20 - 9z = 120.
Combining like terms, we have 4x - 9z = 100.

Therefore, we have found a solution to the equation 4x - 9z = 100, which is (x,z) = (100/4, 0) = (25, 0).
So whenever (x,y) is in R and (y,z) is in R, (x,z) is also in R.

Therefore, R is reflexive, symmetric, and transitive.