1. **State the problem:** We need to find the coordinates of point $P$ given by the vector equation $$P = 5a - 2b$$ where $$a = \begin{pmatrix}3 \\ 2\end{pmatrix}$$ and $$b = \begin{pmatrix}4 \\ 1\end{pmatrix}$$. 2. **Formula and rules:** To find $P$, multiply each vector by its scalar and then subtract the resulting vectors component-wise. 3. **Calculate scalar multiples:** $$5a = 5 \times \begin{pmatrix}3 \\ 2\end{pmatrix} = \begin{pmatrix}5 \times 3 \\ 5 \times 2\end{pmatrix} = \begin{pmatrix}15 \\ 10\end{pmatrix}$$ $$2b = 2 \times \begin{pmatrix}4 \\ 1\end{pmatrix} = \begin{pmatrix}2 \times 4 \\ 2 \times 1\end{pmatrix} = \begin{pmatrix}8 \\ 2\end{pmatrix}$$ 4. **Subtract vectors:** $$P = 5a - 2b = \begin{pmatrix}15 \\ 10\end{pmatrix} - \begin{pmatrix}8 \\ 2\end{pmatrix} = \begin{pmatrix}15 - 8 \\ 10 - 2\end{pmatrix} = \begin{pmatrix}7 \\ 8\end{pmatrix}$$ 5. **Final answer:** The coordinates of point $P$ are $$\boxed{(7, 8)}$$.