Two sides of a triangle are represented by the vectors m=(2/3) and n=(3/2). If r=pm+qn where r=(-1/5), find the values of the scalars p and q.

We are given $m = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$, $n = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$, and $r = \begin{pmatrix} 1 \\ -5 \end{pmatrix}$.

$$\begin{pmatrix} 1 \\ -5 \end{pmatrix} = p \begin{pmatrix} 2 \\ 3 \end{pmatrix} + q \begin{pmatrix} 3 \\ 2 \end{pmatrix}$$

From the top components:

$$2p + 3q = 1 \quad (1)$$

From the bottom components:

$$3p + 2q = -5 \quad (2)$$

Multiply equation (1) by 3 and equation (2) by 2 to eliminate $p$:

$$3 \times (2p + 3q = 1) \implies 6p + 9q = 3 \quad (3)$$ $$2 \times (3p + 2q = -5) \implies 6p + 4q = -10 \quad (4)$$

Subtract equation (4) from equation (3):

$$(6p + 9q) - (6p + 4q) = 3 - (-10)$$ $$5q = 13$$ $$q = \frac{13}{5}$$

Substitute the value of $q$ into equation (1):

$$2p + 3\left(\frac{13}{5}\right) = 1$$ $$2p + \frac{39}{5} = 1$$ $$2p = 1 - \frac{39}{5}$$ $$2p = \frac{5}{5} - \frac{39}{5}$$ $$2p = -\frac{34}{5}$$ $$p = \frac{-34}{5 \times 2}$$ $$p = -\frac{34}{10}$$ $$p = -\frac{17}{5}$$

The values of the scalars are $p = -\frac{17}{5}$ and $q = \frac{13}{5}$.

The final answer is $p = -\frac{17}{5}, q = \frac{13}{5}$.