The problem asks us to find the equation of a line in the form $y = mx + b$ that passes through the points $(-3, 1)$ and $(6, 4)$. We need to find the slope $m$ and the y-intercept $b$.

1. Problem Description

The problem asks us to find the equation of a line in the form $y = mx + b$ that passes through the points $(-3, 1)$ and $(6, 4)$. We need to find the slope $m$ and the y-intercept $b$.

2. Solution Steps

First, we calculate the slope $m$ using the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Given the points $(-3, 1)$ and $(6, 4)$, we have $x_1 = -3$, $y_1 = 1$, $x_2 = 6$, and $y_2 = 4$.

$m = \frac{4 - 1}{6 - (-3)} = \frac{3}{6 + 3} = \frac{3}{9} = \frac{1}{3}$

Now that we have the slope $m = \frac{1}{3}$, we can use the point-slope form of a line:

$y - y_1 = m(x - x_1)$

Using the point $(-3, 1)$ and the slope $m = \frac{1}{3}$, we get:

$y - 1 = \frac{1}{3}(x - (-3))$

$y - 1 = \frac{1}{3}(x + 3)$

$y - 1 = \frac{1}{3}x + 1$

$y = \frac{1}{3}x + 1 + 1$

$y = \frac{1}{3}x + 2$

Thus, the equation of the line is $y = \frac{1}{3}x + 2$. Here $m = \frac{1}{3}$ and $b = 2$.

3. Final Answer

$y = \frac{1}{3}x + 2$