1. **State the problem:** Given vectors $x = \begin{bmatrix}2 \\ 3\end{bmatrix}$, $y = \begin{bmatrix}5 \\ -2\end{bmatrix}$, and $z = \begin{bmatrix}-4 \\ 13\end{bmatrix}$, find scalars $p$ and $q$ such that $$p x + q y = z.$$ 2. **Write the vector equation in component form:** $$p \begin{bmatrix}2 \\ 3\end{bmatrix} + q \begin{bmatrix}5 \\ -2\end{bmatrix} = \begin{bmatrix}-4 \\ 13\end{bmatrix}$$ which gives the system of equations: $$2p + 5q = -4$$ $$3p - 2q = 13.$$ 3. **Solve the system of linear equations:** From the first equation: $$2p = -4 - 5q \implies p = \frac{-4 - 5q}{2}.$$ Substitute $p$ into the second equation: $$3 \left(\frac{-4 - 5q}{2}\right) - 2q = 13.$$ Multiply both sides by 2 to clear the denominator: $$3(-4 - 5q) - 4q = 26.$$ Simplify: $$-12 - 15q - 4q = 26 \implies -12 - 19q = 26.$$ Add 12 to both sides: $$-19q = 38 \implies q = \frac{38}{-19} = -2.$$ 4. **Find $p$ using $q = -2$:** $$p = \frac{-4 - 5(-2)}{2} = \frac{-4 + 10}{2} = \frac{6}{2} = 3.$$ 5. **Final answer:** $$p = 3, \quad q = -2.$$ This means the vector $z$ can be expressed as $3x - 2y$.