Question 174801
<pre><font size = 4 color = "indigo"><b>
Compare 
{{{y=2/3}}}{{{x+1}}} 

with

{{{y=mx+b}}}

We see that the {{{b}}} is {{{1}}}, so that makes
the y-intercept {{{matrix(1,1,"(0,b)")}}} or {{{matrix(1,1,"(0,1)")}}}
and that the slope, {{{m}}}, is {{{2/3}}}.

First we plot the y-intercept {{{matrix(1,1,"(0,1,)")}}}:

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5), locate(-.1,1.2,"@"), locate(-.1,1.2,W),locate(-.1,1.2,O),locate(-.1,1.2,T) )}}}

Now the slope is {{{(rise)/(run)}}}, so we
take the numerator of the slope {{{2/3}}}, which
is {{{2}}}, as the rise, and the denominator,
which is {{{3}}}, as the run.

So we start at the y intercept and draw the rise
of {{{2}}} units upward, indicated below as an upward
pointing arrow.  (If the slope had been negative we 
would have drawn it pointing downward):

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5), locate(-.1,1.2,"@"), locate(-.1,1.2,W),locate(-.1,1.2,O),locate(-.1,1.2,T), line(0,1,0,3),
line(-.1,2.8,0,3),line(.1,2.8,0,3)
 )}}}

Then we go right {{{3}}} units for the run,
(run is always to the right) indicated by the second arrow
pointing to the right:

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5), locate(-.1,1.2,"@"), locate(-.1,1.2,W),locate(-.1,1.2,O),locate(-.1,1.2,T), line(0,1,0,3),
line(-.1,2.8,0,3),line(.1,2.8,0,3), line(0,3,3,3),
line(2.9,2.9,3,3),line(2.9,3.1,3,3) 
 )}}}

Now to graph the line, all we do is draw a
straight line through the y-intercept (0,1)
and through the arrow head at the end of
the run.  So we have this graph:

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5,(2/3)x+1), locate(-.1,1.2,"@"), locate(-.1,1.2,W),locate(-.1,1.2,O),locate(-.1,1.2,T), line(0,1,0,3),
line(-.1,2.8,0,3),line(.1,2.8,0,3), line(0,3,3,3),
line(2.9,2.9,3,3),line(2.9,3.1,3,3) 
 )}}}

Edwin</pre>