I'm having a hard time understanding this portion of my algebra homework. I just can't grasp the concept.

Find the equation of the line parallel to y=3x-2 and passing through (3,2). Put your equation in slope-intercept form.

Look up your slope-intercept form in your book. y=mx+b, where m is slope and b is the y intercept. So you have your slope already given to you.

If two lines are parallel they are going to have the same slope, because they're running side-by-side at the same angle. So your second line must have the same slope as the line you're given.

If you want to write an equation for a line, it has to have the variables x and y in it. If it had actual numbers, it would be a point, not a whole line. So you need a way to get an equation using what you have, and still have those variables in it. You've been given one point of that second line. And you have the slope.

An equation that takes such information would be the point-slope equation:


Plug in what you already have into this equation. (It doesn't matter if your point is sub-0 or 1... just be consistent.) Then you want to solve for y in terms of x. That is, re-arrange everything so that it looks like the slope-intersect equation y=mx+b. The m & b will be numbers, and x and y are left as variables. Use your algebra rules to distribute, move things to the other side, etc, until you get it into that form.

Give that a try.