...(16xy)/(xy) (1)/(y) + (2)/(x) = 16 (4)

Substitute these into equations (3) and (4) to form a linear system.
The system becomes:

$$2b - a = 2 \quad (3')$$ $$b + 2a = 16 \quad (4')$$

From equation (3'), we can express $a$ in terms of $b$:

$$a = 2b - 2$$

Substitute this expression for $a$ into equation (4'):

$$b + 2(2b - 2) = 16$$ $$b + 4b - 4 = 16$$ $$5b - 4 = 16$$ $$5b = 20$$ $$b = 4$$

Now substitute the value of $b$ back into the expression for $a$:

$$a = 2(4) - 2$$ $$a = 8 - 2$$ $$a = 6$$

Since $a = 6$, we have $\frac{1}{x} = 6$, which means $x = \frac{1}{6}$.
Since $b = 4$, we have $\frac{1}{y} = 4$, which means $y = \frac{1}{4}$.

The solution to the system is:

$$x = \frac{1}{6}, y = \frac{1}{4}$$

We can verify these values in the original equations.
For $2x - y = 2xy$: $2\left(\frac{1}{6}\right) - \frac{1}{4} = \frac{1}{3} - \frac{1}{4} = \frac{4-3}{12} = \frac{1}{12}$. And $2\left(\frac{1}{6}\right)\left(\frac{1}{4}\right) = 2\left(\frac{1}{24}\right) = \frac{1}{12}$. This is correct.
For $x + 2y = 16xy$: $\frac{1}{6} + 2\left(\frac{1}{4}\right) = \frac{1}{6} + \frac{1}{2} = \frac{1+3}{6} = \frac{4}{6} = \frac{2}{3}$. And $16\left(\frac{1}{6}\right)\left(\frac{1}{4}\right) = 16\left(\frac{1}{24}\right) = \frac{16}{24} = \frac{2}{3}$. This is also correct.