### Instruction: 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 $k$.### Response: I recognize this equation as a system of linear equations in parametric form, where each vector represents a point or a direction.
For the system to have infinitely many solutions, the direction vectors must be parallel, which means they are proportional to each other.
So I can write \[\begin{pmatrix} 5 \\ 2 \end{pmatrix} = \lambda \begin{pmatrix} -15 \\ -6 \end{pmatrix}\] for some constant $\lambda$.
This implies that $5 = -15 \lambda$ and $2 = -6 \lambda$, which means that $\lambda = -\frac{1}{3}$.
Therefore, the direction vectors are equal to \[\begin{pmatrix} 5 \\ 2 \end{pmatrix} = -\frac{1}{3} \begin{pmatrix} -15 \\ -6 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}.\]
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 \[k = \frac{8}{5} - 1 = \frac{3}{5}.\]