Given the relation $R = \{(x, y) : 4y = 5x - 3, x, y \in M\}$, where $M = \{1, 2, 3, ..., 16\}$.

First, we need to find the elements in $R$. We can rewrite the equation as $y = \frac{5x - 3}{4}$. Since $x, y \in M$, we need to find integer values of $x$ such that $1 \le x \le 16$ and $1 \le y \le 16$.

We can test values of $x$ from 1 to 16:

• If $x = 1$, $y = \frac{5(1) - 3}{4} = \frac{2}{4} = 0.5$, which is not in $M$.
• If $x = 3$, $y = \frac{5(3) - 3}{4} = \frac{12}{4} = 3$, so $(3, 3) \in R$.
• If $x = 7$, $y = \frac{5(7) - 3}{4} = \frac{32}{4} = 8$, so $(7, 8) \in R$.
• If $x = 11$, $y = \frac{5(11) - 3}{4} = \frac{52}{4} = 13$, so $(11, 13) \in R$.
• If $x = 15$, $y = \frac{5(15) - 3}{4} = \frac{72}{4} = 18$, which is not in $M$.

Thus, $R = \{(3, 3), (7, 8), (11, 13)\}$.

For $R$ to be symmetric, if $(x, y) \in R$, then $(y, x)$ must also be in $R$.

Currently, $(3, 3) \in R$, and since $x = y$, this element doesn't require any additional element for symmetry.

However, $(7, 8) \in R$, but $(8, 7) \notin R$. We need to add $(8, 7)$ to $R$ to make it symmetric.

Also, $(11, 13) \in R$, but $(13, 11) \notin R$. We need to add $(13, 11)$ to $R$ to make it symmetric.

Therefore, we need to add $(8, 7)$ and $(13, 11)$ to $R$ to make it symmetric. The number of elements to be added is 2.