How do you graph #2x+3y=3# using intercepts?

1 Answer
Jul 16, 2017

Answer:

Please see below.

Explanation:

#x#-intercept is where #y=0# i.e. #2x+3xx0=3# or #2x=3# and #x=1.5# i.e. line passes through #(1.5,0)#.

#y#-intercept is where #x=0# i.e. #2xx0+3y=3# or #3y=3# and #y=1# i.e. line passes through #(0,1)#.

Now join #(1.5,0)# and #(0,1)# to get the line.

graph{(2x+3y-3)((x-1.5)^2+y^2-0.003)(x^2+(y-1)^2-0.003)=0 [-1.917, 3.082, -0.77, 1.73]}

Tip - If intercept on #x#-axis is#a# and intercept on #y#-axis is #b#, equation of line is #x/a+y/b=1#.