How do you graph using slope and intercept of #3/2x - 1/3y = 7/3#?

1 Answer
Dec 31, 2015

Manipulate so that the equation is in the form #y=mx+b#, where #m# is the slope and #b# is the #y#-intercept.

First, to isolate the #y#-term, subtract #3/2x# from both sides.

#-1/3y=-3/2x+7/3#

To isolate just #y#, multiply both sides by #-3#.

#y=9/2x-7#

This means that the function has an intercept at #(0,-7)#.

The slope is #9/2#. Slope is also known as "rise over run", since it represents the change in #y#-values over the change in #x#-values.

Thus, the "rise" is #9# for every "run" of #2#. This means that if you have the point #(0,-7)#, you know that the point #(0+2,-7+9)=>(2,2)# is also on the line.

Graph #(2,2)# and draw the line. If you wish, you can continue going up #9# units and right #2# units to draw in more points. You can also go down #9# units and left #2# units.

graph{3/2x-1/3y=7/3 [-28.47, 44.6, -15.8, 20.72]}