We have that $x = 8 + 2t$ and $y = -1 + 3t.$  Isolating $t$ in $x = 8 + 2t,$ we find
\[t = \frac{x - 8}{2}.\]Then
\begin{align*}
y &= -1 + 3t \\
&= -1 + 3 \cdot \frac{x - 8}{2} \\
&= \frac{3}{2} x - 13.
\end{align*}Thus, $(m,b) = \boxed{\left( \frac{3}{2}, -13 \right)}.$
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
A line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \end{pmatrix}.\]The equation of the line can be expressed in the form $y = mx + b.$  Enter the ordered pair $(m,b).$
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
\left( \frac{3}{2}, -13 \right)
A line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \end{pmatrix}.\]The equation of the line can be expressed in the form $y = mx + b.$  Enter the ordered pair $(m,b).$
test