There exists a real number $k$ such that the equation
\[\begin{pmatrix} 4 \\ -1 \end{pmatrix} + t \begin{pmatrix} 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ k \end{pmatrix} + s \begin{pmatrix} -15 \\ -6 \end{pmatrix}\]has infinitely many solutions in $t$ and $s$.  Find $k$.
This implies that $5 = -15 \lambda$ and $2 = -6 \lambda$, which means that $\lambda = -\frac{1}{3}$. This also means that the points must lie on the same line, which is determined by either of the direction vectors. I can use the point-slope form of a line to write the equation of the line as \[y - (-1) = \frac{2}{5}(x - 4),\] using the point $(4, -1)$ and the slope $\frac{2}{5}$. To find the value of $k$, I need to plug in the point $(8, k)$ into this equation and solve for $k$. So I get \[k - (-1) = \frac{2}{5}(8 - 4),\] which simplifies to \[k + 1 = \frac{8}{5}.\] Subtracting 1 from both sides, I get \[k = \frac{8}{5} - 1 = \frac{3}{5}.\] 