Solve the following: $$\begin{cases} 4 - 4\frac{1}{3} \\ \begin{cases} 5p - 3q = 0 \\ 3p + 4q = 29 \end{cases} \end{cases}$$

Let's solve this in two parts: Part 1: Simple Arithmetic $$4 - 4\frac{1}{3} = 4 - \frac{13}{3} = \frac{12}{3} - \frac{13}{3} = -\frac{1}{3}$$ Part 2: System of Equations We have the following system of equations: $$\begin{cases} 5p - 3q = 0 \\ 3p + 4q = 29 \end{cases}$$ We can solve this system using substitution or elimination. Let's use substitution. From the first equation, we can express $$p$$ in terms of $$q$$: $$5p = 3q \Rightarrow p = \frac{3}{5}q$$ Now substitute this expression for $$p$$ into the second equation: $$3\left(\frac{3}{5}q\right) + 4q = 29$$ $$\frac{9}{5}q + 4q = 29$$ $$\frac{9}{5}q + \frac{20}{5}q = 29$$ $$\frac{29}{5}q = 29$$ $$q = 29 \cdot \frac{5}{29} = 5$$ Now that we have $$q$$, we can find $$p$$: $$p = \frac{3}{5}q = \frac{3}{5}(5) = 3$$ So the solution to the system of equations is $$p = 3$$ and $$q = 5$$. Final Answer: For the arithmetic part: $$\boxed{-\frac{1}{3}}$$ For the system of equations: $$\boxed{p=3, q=5}$$