Use the slope and y-intercept to graph the equation: 3y+15x=0

To see the slope and y-intercept more clearly, get y by itself. 
3y=-15x
y=(-15/3)x
y=-5x
If you use the form y=mx+b, you can see that the slope is -5, and the y-intercept is 0. This means you will have a point at (0,0), and then you will go down 5, and across to the right 1, and so on.

Slope int form; y=mx+b

first to clearly see slope and y-int you must get equation into slope int form;
3y+15x=0
3y=-15x
y=-15/3x
y=-5x
y=-5x+0

so you're slope (m) is equal to " -5 " , and your y-int (b) is equal to " 0 "

so in order to graph this you would have to start at your y-int point which is (0,0) because in this case you don't have a y-int. from there you you can use the " rise and run " method with your slope ( so basically in this case you can go up or down -5 and then across one and then up or down again 5 if that makes sense )

here are a few plot points from this equations graph to check yourself; 
(1,-8)
(1 ,8)
(0,0)
(2,-16)

hope this helped good luck c: