First, let's rewrite this as point-slope form or #y=mx+b#:
#x-2/3y=1#
subtract #x# on both sides
#-2/3y=1-x#
divide both sides by #-2/3#
#y=(1-x) -: -2/3#
#y= (1-x)/1 xx -3/2#
#y= (2(-x+1))/3#
#y=(-2x+2)/3#
#y=-(2)/3x + 2/3#
Now it's in point-slope form ( #y=color(red)(m)x+color(blue)(b)# )
the reason I wanted it in this exact form is because it gives us valuable information: the #color(red)(slope)# and the #color(blue)(y-i ntercept)#
These are very helpful for graphing, buecause we really only need to points on the line to draw the rest of it. And we already have one, the #y#-intercept:
#y=color(red)(-2/3)x+color(blue)(2/3)#
The #y#-intercept is #(0, 2/3)#, and it's very simple to find the #x#-intercept. All we need to do is set #y# equal to zero and solve for #x#:
#y=-(2)/3x + 2/3#
#0=-(2)/3x + 2/3#
subtract #2/3# on both sides
#-2/3=-2/3x#
divide by #-2/3# on both sides
#-2/3 -: -2/3 = x#
#cancel-cancel2/cancel3 xx cancel-cancel3/cancel2 = x#
#x=1#
So, now we have two pints: #(0, 2/3)# and #(1, 0)#. Now you just need to use a rule to draw a line between these points, or start at one of them and use the slope to find more points:
Just to check our work, let's graph our equation. If we did everything correctly, the line should pass through the points #(0, 2/3)# and #(1, 0)#.
graph{y=-(2)/3x + 2/3}
We were right!
