When you plug $y = \dfrac{2a+3}{a-4}$ into $\dfrac{2y+3}{y-4}$, carefully combine fractions so that the numerator and denominator each become a single fraction in terms of $a$, then simplify the overall expression.

Once you have a simplified expression for $f(f(a))$ in terms of $a$, set it equal to $a$ and clear the denominator by multiplying both sides by that denominator. Rearrange the resulting equation into the standard quadratic form $a^2 + ba + c = 0$.

After you have the quadratic equation, you do not need the individual solutions to find the sum of the elements of $S$. Use the fact that for $x^2 + bx + c = 0$, the sum of the solutions is $-b$.

In the first expression line, type f(x) = (2x+3)/(x-4) so Desmos defines the function $f$.

In the next line, type g(x) = (2*f(x)+3)/(f(x)-4) so that $g(x)$ represents $f(f(x))$.

Add another line with y = x. Look at the graph of $y = g(x)$ and $y = x$ and find their intersection points. The $x$-coordinates of these intersection points are the values of $a$ that satisfy $f(f(a)) = a$ (ignore the vertical asymptotes near $x = 4$ and $x = 9.5$).

Record the two $x$-values from the intersection points and add them together (you can type their approximate values in a new expression like a1 + a2). The result of this addition should match one of the answer choices.

We are given

for all real $x \neq 4$. We want all real numbers $a$ such that

So our goal is to:

• Find a formula for $f(f(x))$.
• Set $f(f(a)) = a$ and solve for $a$, remembering that $a \neq 4$ and also $f(a) \neq 4$ (so the inner $f$ is defined).

Let $y=f(x)$. Then

Now

Replace $y$ with $\dfrac{2x+3}{x-4}$:

• Numerator:

• Denominator:

So

with the understanding that $x \neq 4$ and $x \neq \dfrac{19}{2}$ (since $19-2x \neq 0$).

Now apply the composition to $a$:

We are given $f(f(a)) = a$, so

with $a \neq 4$ and $a \neq \dfrac{19}{2}$.

Clear the fraction by multiplying both sides by $19-2a$ (which we know is not zero for valid $a$):

Expand the right side and bring all terms to one side:

Multiply by $-1$ and divide by $2$ to simplify:

So any valid $a$ satisfying $f(f(a))=a$ must be a solution to this quadratic equation (and must not be $4$ or $\dfrac{19}{2}$).

The equation

is a monic quadratic of the form $a^2 + ba + c = 0$ with $b = -6$ and $c = -3$.

For any quadratic $x^2 + bx + c = 0$, the sum of its solutions is $-b$.

Here, $-b = -(-6)$, so the sum of the solutions to $a^2 - 6a - 3 = 0$ is

Both solutions are real and are not equal to $4$ or $\dfrac{19}{2}$, so they are both valid elements of $S$. Therefore, the sum of the elements of $S$ is $\boxed{6}$, which corresponds to answer choice C.