The given task asks to write an equation of a line parallel to $y = \frac{7}{8}x + 1$ that passes through the point $C(3,5)$.

To solve this:

1. Since the line is parallel to $y = \frac{7}{8}x + 1$, it will have the same slope. The slope of this line is $\frac{7}{8}$.

2. The general equation for a line in slope-intercept form is: $y = mx + b$ where $m$ is the slope and $b$ is the y-intercept.

3. To find the equation of the new line passing through $C(3,5)$, we use the point-slope form of a line: $y - y_1 = m(x - x_1)$ where $(x_1, y_1) = (3, 5)$ and $m = \frac{7}{8}$. Substituting the values: $y - 5 = \frac{7}{8}(x - 3)$

4. Simplify the equation: $y - 5 = \frac{7}{8}x - \frac{21}{8}$ $y = \frac{7}{8}x - \frac{21}{8} + 5$ $y = \frac{7}{8}x - \frac{21}{8} + \frac{40}{8}$ $y = \frac{7}{8}x + \frac{19}{8}$

Thus, the equation of the line parallel to $y = \frac{7}{8}x + 1$ that passes through $C(3,5)$ is: $y = \frac{7}{8}x + \frac{19}{8}$

Would you like any more details or have questions about this?

Here are 5 related questions you might explore:

1. How do you find the slope of a line given two points?
2. What are the different forms of linear equations?
3. How do you convert a line equation from point-slope to slope-intercept form?
4. What is the significance of the y-intercept in the slope-intercept form?
5. How would the equation change if the line were perpendicular instead of parallel?

Tip: The slope of parallel lines is always the same, while perpendicular lines have slopes that are negative reciprocals.