First, let's rewrite this as point-slope form or y=mx+by=mx+b:
x-2/3y=1x−23y=1
subtract xx on both sides
-2/3y=1-x−23y=1−x
divide both sides by -2/3−23
y=(1-x) -: -2/3y=(1−x)÷−23
y= (1-x)/1 xx -3/2y=1−x1×−32
y= (2(-x+1))/3y=2(−x+1)3
y=(-2x+2)/3y=−2x+23
y=-(2)/3x + 2/3y=−23x+23
Now it's in point-slope form ( y=color(red)(m)x+color(blue)(b)y=mx+b )
the reason I wanted it in this exact form is because it gives us valuable information: the color(red)(slope)slope and the color(blue)(y-i ntercept)y−intercept
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 yy-intercept:
y=color(red)(-2/3)x+color(blue)(2/3)y=−23x+23
The yy-intercept is (0, 2/3)(0,23), and it's very simple to find the xx-intercept. All we need to do is set yy equal to zero and solve for xx:
y=-(2)/3x + 2/3y=−23x+23
0=-(2)/3x + 2/30=−23x+23
subtract 2/323 on both sides
-2/3=-2/3x−23=−23x
divide by -2/3−23 on both sides
-2/3 -: -2/3 = x−23÷−23=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!
