For part a, we need to get the equation to be #y=mx+b#, so we do the following:
#5x-3y=15#
#5x-3ycolor(red)(+3y)=15color(red)(+3y)#
#5xcolor(red)(-15)=15+3ycolor(red)(-15)#
#(5x-15)/color(red)(3)=(3y)/color(red)(3)#
#5/3x-5=y#
For part b, we know that in the equation #y=mx+b# #m# is the slope, so from our equation in a we know that #5/3# is the slope.
For part c, we know that in the equation #y=mx+b# the y-intercept is at #(0,b)#, so our point is at #(0,-5)#
For part d, we combine parts c and b to know that we start at #(0,-5)# and go up 5 and across 3 (slope of 5/3). Another way to graph it is by taking the equation we were given and setting #x=0# and finding #y# and then setting #y=0# and finding x, those are your intercepts and can be graphed to create a line equal to the one you want.
Hope this helped!
