Find the value of $k$ for which the line passing through the points $(3, 2k+1)$ and $(8, 4k-5)$ is parallel to the x-axis.

1. Problem Description

Find the value of $k$ for which the line passing through the points $(3, 2k+1)$ and $(8, 4k-5)$ is parallel to the x-axis.

2. Solution Steps

A line is parallel to the x-axis if its slope is

0. The slope $m$ of the line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

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

In our case, $(x_1, y_1) = (3, 2k+1)$ and $(x_2, y_2) = (8, 4k-5)$. Therefore, the slope is:

$m = \frac{(4k - 5) - (2k + 1)}{8 - 3}$

Since the line is parallel to the x-axis, we set the slope to 0:

$0 = \frac{4k - 5 - 2k - 1}{5}$

$0 = \frac{2k - 6}{5}$

Multiplying both sides by 5:

$0 = 2k - 6$

$2k = 6$

$k = \frac{6}{2}$

$k = 3$

3. Final Answer

The value of $k$ is 3.