Twice the sum of the digits of a positive, two-digit integer is 19 greater than the ones digit. Reversing the digits decreases the number by 18.

If the number is ab, then we have

2(a+b) = b+19
10b+a = 10a+b - 18

Now solve for a and b, and you see that the number is 75

To solve this problem, let's assume that the two-digit integer is represented by the variables 't' and 'u', where 'u' represents the ones digit and 't' represents the tens digit.

Based on the given information, we can form two equations:

Equation 1: Twice the sum of the digits of the number is 19 greater than the ones digit.
2(t+u) = u + 19

Equation 2: Reversing the digits decreases the number by 18.
10u + t - 18 = 10t + u

Now, let's solve these equations step by step:

First, simplify Equation 1:
2t + 2u = u + 19

Next, simplify Equation 2:
9u - 9t = 18

To make the equations easier to solve, let's isolate 'u' in Equation 1:
2t + u = 19

Now, we can solve this system of equations algebraically:

Step 1: Solve Equation 1 for 't':
t = (19 - u)/2

Step 2: Substitute this expression for 't' into Equation 2:
9u - 9((19 - u)/2) = 18

Simplify the equation:
9u - 9(19 - u)/2 = 18

Multiply both sides by 2 to eliminate the denominator:
18u - 9(19 - u) = 36

Distribute:
18u - 171 + 9u = 36

Combine like terms:
27u - 171 = 36

Add 171 to both sides:
27u = 207

Divide by 27:
u = 7

Now that we have the value for 'u', substitute it into Equation 1 to find 't':
2t + 7 = 19

Subtract 7 from both sides:
2t = 12

Divide by 2:
t = 6

So, the original two-digit number is 67.

To summarize, the original two-digit integer is 67.