Question #763da

1 Answer
Feb 1, 2018

Answer:

See below...

Explanation:

Firstly, I would get the line equation in the

#y=mx+c# format.

#therefore 3y= 2x-6#

#y = (2/3)x - 2#

Now I would make a table of values ranging from the #x# values #-3# to #3#

Each #x# value corresponds to a #y# value giving you a set of coordinates.

E.g when #x =3#, #y=(2/3)(3)-2=0#
#therefore (3,0)#

After getting these coordinates, plot these points on a four quadrant axis and finish with a smooth, straight line. graph{-2x+3y=-6 [-10, 10, -5, 5]}