How do you find the equation of a line parallel to y = 3x + 5 through (1, 2)?

The equation of a line parallel to \( y = 3x + 5 \) through \( (1, 2) \) is \( y = 3x - 1 \).

To find the equation of a line parallel to a given line, we need to understand that parallel lines have the same gradient (slope). The given line \( y = 3x + 5 \) has a gradient of 3. Therefore, any line parallel to this one will also have a gradient of 3.

Next, we use the point-slope form of the equation of a line, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the gradient and \( (x_1, y_1) \) is a point on the line. Here, \( m = 3 \) and the point given is \( (1, 2) \).

Substituting these values into the point-slope form, we get:
\[ y - 2 = 3(x - 1) \]

Now, we simplify this equation to get it into the slope-intercept form \( y = mx + c \):
\[ y - 2 = 3x - 3 \]
\[ y = 3x - 3 + 2 \]
\[ y = 3x - 1 \]

So, the equation of the line parallel to \( y = 3x + 5 \) that passes through the point \( (1, 2) \) is \( y = 3x - 1 \).