The problem asks to find the equation of a line that is parallel to the given line and passes through the point $(3, -1)$.

Step 1: Find the slope of the given line

The given line is in standard form:
$8x - 3y = 5$

We need to rewrite this in slope-intercept form $y = mx + b$ to find the slope.

First, solve for $y$: $-3y = -8x + 5$ $y = \frac{8}{3}x - \frac{5}{3}$ The slope $m$ of this line is $\frac{8}{3}$.

Step 2: Write the equation of the parallel line

Parallel lines have the same slope, so the slope of the new line is also $\frac{8}{3}$.

We use the point-slope form to find the equation of the line passing through $(3, -1)$: $y - y_1 = m(x - x_1)$ where $m = \frac{8}{3}$, $(x_1, y_1) = (3, -1)$. Substituting the values: $y - (-1) = \frac{8}{3}(x - 3)$ $y + 1 = \frac{8}{3}(x - 3)$

Step 3: Simplify the equation

Distribute the slope on the right side: $y + 1 = \frac{8}{3}x - 8$ Subtract 1 from both sides: $y = \frac{8}{3}x - 9$

Thus, the equation of the line parallel to $8x - 3y = 5$ and passing through $(3, -1)$ is: $y = \frac{8}{3}x - 9$

Let me know if you'd like more details or have any questions!

Here are 5 related questions for practice:

1. What is the equation of a line parallel to $2x - y = 4$ and passing through $(5, 2)$?
2. How would the equation change if the new line was perpendicular to the original line?
3. Find the slope of a line perpendicular to the line $4x + 5y = 10$.
4. How do you convert an equation from standard form to slope-intercept form?
5. What is the equation of a line passing through $(-2, 3)$ and parallel to $y = -\frac{1}{2}x + 6$?

Tip:

For parallel lines, always remember that they share the same slope. For perpendicular lines, their slopes are negative reciprocals of each other.