Bobby's concern about the equation being unsolvable due to a variable in the denominator is valid, but we can address it by finding the restricted value of \(x\) that makes the denominator zero. Let's analyze the equation:
\[ \frac{3}{x} + \frac{1}{3} = \frac{5}{6} \]
To find the solution, first, find a common denominator, which is \(3x\) in this case. Multiply each term by \(3x\):
\[ 3 + x = \frac{5x}{6} \]
Now, multiply both sides of the equation by 6 to eliminate the fraction:
\[ 18 + 6x = 5x \]
Subtract \(5x\) from both sides:
\[ 18 = -x \]
Multiply both sides by -1 to solve for \(x\):
\[ x = -18 \]
Now, it's true that when \(x = -18\), the denominator \(x\) becomes zero, but this is a restricted value that we need to consider. So, the equation is solvable, but we should note that \(x = -18\) is not in the domain of the original equation due to the division by zero issue.