How do you graph #x+3y=6# by plotting points?

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Answer 1 · 2017-04-02T03:51:32

A straight line through #(0,2)# and #(6,0)#

The equation of a straight line in slope #(m)# and intercept #(c)# form is:

#y=mx+c#

Rewriting our equation in this form #-> y=-x/3+2#

Hence the equation is a straight line with slope #-1/3# and y-intercept #+2#

We are asked to graph the line by plotting points.

For simplicity, I will choose the points where #x=0# and where #y=0# (NB: Any two points on this line would do)

#x=0 -> y=2#

#y=0 -> -x/3+2=0 -> x=6#

Hence the graph is a straight line through #(0,2)# and #(6,0)# as shown below:

graph{-x/3+2 [-2.446, 10.04, -2.695, 3.55]}