<Question>
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 $(m,b).$
</Question>

<think>
<Plan_1>
Okay, so I need to find the equation of the line in the form y = mx + b, right? The line is given in parametric form. Let me recall how parametric equations relate to the slope-intercept form.

<prompt_1.1> Recall how parametric equations relate to the slope-intercept form and identify the components of the parametric equation. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The parametric equations for a line are usually given as x = x0 + at and y = y0 + bt, where (x0, y0) is a point on the line and the direction vector is (a, b). In this case, the parametric equations are:

x = 8 + 2t

y = -1 + 3t

So, the point (8, -1) is on the line, and the direction vector is (2, 3). The slope of the line should be the ratio of the change in y over the change in x, which is b/a. Here, that would be 3/2. So, m should be 3/2.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, to find b, the y-intercept, I need to find where the line crosses the y-axis, which is when x = 0. But maybe there's a better way than plugging in x = 0. Since I already have a point (8, -1) that's on the line, I can use the point-slope form.

<prompt_2.1> Use the point-slope form to find the equation of the line. </prompt_2.1>
-----

Alternatively, maybe I can use another method. Let me see. If I solve the parametric equations for t and set them equal, that should give the Cartesian equation.

<prompt_2.2> Solve the parametric equations for t and set them equal to find the Cartesian equation. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
The point-slope formula is y - y1 = m(x - x1). Let's plug in the values. Here, (x1, y1) is (8, -1), and m is 3/2. So:

y - (-1) = (3/2)(x - 8)

Simplify that:

y + 1 = (3/2)x - (3/2)*8

Calculate (3/2)*8: 3*4 = 12. So,

y + 1 = (3/2)x - 12

Subtract 1 from both sides:

y = (3/2)x - 12 - 1

Which is:

y = (3/2)x - 13
</execution_2.1>

<execution_2.2>
From the x equation: x = 8 + 2t. Let's solve for t:

x - 8 = 2t

t = (x - 8)/2

Now plug this into the y equation:

y = -1 + 3t = -1 + 3*(x - 8)/2

Let me simplify that:

y = -1 + (3x - 24)/2

Convert -1 to -2/2 to combine the terms:

y = (-2/2) + (3x - 24)/2 = (3x - 24 - 2)/2 = (3x - 26)/2 = (3/2)x - 13

So again, m is 3/2 and b is -13.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, is that right?

Wait, let me check my steps again.

<prompt_3.1> Verify the steps to ensure the equation is correct. </prompt_3.1>
-----

Based on execution_2.1, execution_2.2:
Let me just verify once more with a different point. Suppose t = 1, then the point is (10, 2).

<prompt_3.2> Verify the equation using a different point. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Starting with:

y - (-1) = (3/2)(x - 8)

So y + 1 = (3/2)x - 12

Subtract 1: y = (3/2)x - 12 - 1

Which is y = (3/2)x - 13.
</execution_3.1>

<execution_3.2>
Plugging into y = (3/2)x -13: (3/2)*10 = 15, 15 -13 = 2. Correct. Another point: t = 0 gives (8, -1). Plugging in: (3/2)*8 = 12, 12 -13 = -1.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Alternatively, I can convert the parametric equations directly into slope-intercept form. Let's see.

<prompt_4.1> Convert parametric equations to slope-intercept form. </prompt_4.1>

Wait, but let me check if I made a mistake in the signs. When I subtract 12 -1, that's -13.

<prompt_4.2> Verify the y-intercept calculation. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
From the parametric equations:

x = 8 + 2t ⇒ solve for t: t = (x - 8)/2

Then substitute that into the y equation:

y = -1 + 3t = -1 + 3*( (x - 8)/2 ) = -1 + (3x - 24)/2

Combine terms:

Convert -1 to -2/2: (-2/2) + (3x -24)/2 = (3x -24 -2)/2 = (3x -26)/2 = (3/2)x -13
</execution_4.1>

<execution_4.2>
Yes. So the y-intercept is indeed -13.
</execution_4.2>
</Execution_4>

<Final_answer>
Therefore, the equation of the line in the form $y = mx + b$ is $y = \frac{3}{2}x - 13$, and the ordered pair $(m,b)$ is $\boxed{\left( \frac{3}{2},-13 \right)}$.
</Final_answer>
</think>