Help with some algebra 1 homework

I have a test tomorrow and can't figure some of this out . . .

1. Write the equation of a line that is parallel to y= -3x+5 and goes through the point (6,-1)

2.Write the equation of a line parallel to the x axis that is units below the x axis

3. if x varies directly as y, and x= 6 when y=22, write the equation that relates x and y

4. solve and graph 2x – 3y < 6.

1. Parallel lines have the same slope. A line parallel to y=-3x+5 would have to also have a slope of -3. You should plug in that slope and your coordinates that the line passes through (6,-1) into the slope intercept form of a linear equation. So you get this equation
-1= -3(6)+b
-1=-18+b
17=b

Then you plug in the slope and y intercept to write the equation
y=-3x+17

2. A line parallel to the x axis is horizontal. Horizontal lines are written as y = ?.
if you want a horizontal line that goes through (2,3) it has to be y = 3 because 3 is the y value of the point. The problem you wrote down left off how many units below the x axis it needs to be. But if it's 5 units below x it would be y= -5.

3. In a direct variation is y=kx and k = y/x. So if y= 22 when x=6 then k= 22/6 or 11/3.
Since that is k then you can write the direct variation as y = 11/3 x. K is basically the slope of a direct variation. All direct variation must go through the origin so they have no constants added or subtracted in the equation.

4. The graph of 2x -3y < 6 is a shaded region of the graph bellow the line. Since the inequality sign is < connect the intercepts with a dashed line and you should shade below the line. Remember that to find the intercepts just plug in 0 for x and y.

1. If the line is parallel, that means that it's going to have the same slope as the one you already have, so -3.
Now, you know that your second line must go through (6, -1). You only lack b, let's find it: