Question 480989
a) y=3x+1
b) 3x-4y=12
<pre>
I'll do the second one first, since it requires more work,
then I'll do the first one:

Solve it for y:

3x - 4y = 12
    -4y = -3x + 12

Divide through by 4

    {{{(-4)/(-4)}}}y = {{{(-3)/(-4)}}}x + {{{12/(-4)}}}

          y = {{{3/4}}}x - 3

compare that to 

          y = mx + b

the slope m is {{{3/4}}} and the y-intercept is (0,b) or (0,-3)

We begin at the y-intercept (0,-3):

{{{drawing(400,400,-6,6,-6,6, graph(400,400,-6,6,-6,6),

locate(0,-3,"(0,-3)"), circle(0,-3,.1)  )}}}
         
We take the numerator of the slope which is 3, and since the
slope is positive we draw a line from the y-intercept UPWARD
3 units, like the green arrow below. [If the slope had been
negative, which it wasn't, we would have draw the arrow downward]

{{{drawing(400,400,-6,6,-6,6, graph(400,400,-6,6,-6,6),
green(line(0,-3,0,0), line(-.2,-.3,0,0),line(.2,-.3,0,0)),
locate(0,-3,"(0,-3)"), circle(0,-3,.1)  )}}}

Now we take the denominator of the slope which is 4, and draw a 
another line from that green arrow head to the RIGHT 4 units, as 
shown below. [Even if the slope had been negative, we still would have 
drawn the arrow to the RIGHT the number of units indicated by the
denominator of the slope.]


{{{drawing(400,400,-6,6,-6,6, graph(400,400,-6,6,-6,6),
green(line(0,-3,0,0), line(-.2,-.3,0,0),line(.2,-.3,0,0), line(0,0,4,0),

line(3.7,.2,4,0),line(3.7,-.2,4,0)),

locate(0,-3,"(0,-3)"), circle(0,-3,.1)  )}}}

Now get a ruler or straight edge and draw a straight line
through the y-intercept (0,-3) and through the second arrow head:

{{{drawing(400,400,-6,6,-6,6, graph(400,400,-6,6,-6,6),
green(line(0,-3,0,0), line(-.2,-.3,0,0),line(.2,-.3,0,0), line(0,0,4,0),

line(3.7,.2,4,0),line(3.7,-.2,4,0)),
line(-12,-12,12,6),
locate(0,-3,"(0,-3)"), circle(0,-3,.1)  )}}}

-----------------------------

Now I'll do the first one:

          y = 3x + 1

It is already solved for y and is in slope-y-intercept form,
so we don't have to do that like we did in the other one:

compare that to 

          y = mx + b

the slope m is 3 and the y-intercept is (0,b) or (0,1)

As before we begin at the y-intercept (0,1):

{{{drawing(400,400,-6,6,-6,6, graph(400,400,-6,6,-6,6),

locate(0,1,"(0,1)"), circle(0,1,.1)  )}}}
        
Now the slope 3 in this one doesn't have a visible denominator,
so we write it as {{{3/1}}} so it will have both a numerator and
a denominator:  

We take the numerator of the slope which is 3, and since the
slope is positive we draw a line from the y-intercept UPWARD
3 units, like the green arrow below. [If the slope had been
negative, which it wasn't, we would have draw the arrow downward]

{{{drawing(400,400,-6,6,-6,6, graph(400,400,-6,6,-6,6),
green(line(0,1,0,4), line(-.2,3.7,0,4),line(.2,3.7,0,4)),
locate(0,1,"(0,1)"), circle(0,1,.1)  )}}}

Now we take the denominator of the slope which is 1, and draw a 
another line from that green arrow head to the RIGHT 1 unit, as 
shown below. [Even if the slope had been negative, we still would have 
drawn the arrow to the RIGHT the number of units indicated by the
denominator of the slope.]


{{{drawing(400,400,-6,6,-6,6, graph(400,400,-6,6,-6,6),
green(line(0,1,0,4), line(-.2,3.7,0,4),line(.2,3.7,0,4), line(0,4,1,4),

line(.7,3.8,1,4),line(.7,4.2,1,4)),

locate(0,1,"(0,1)"), circle(0,1,.1)  )}}}

Now get a ruler or straight edge and draw a straight line
through the y-intercept (0,1) and through the second arrow head:


{{{drawing(400,400,-6,6,-6,6, graph(400,400,-6,6,-6,6),
green(line(0,1,0,4), line(-.2,3.7,0,4),line(.2,3.7,0,4), line(0,4,1,4),
line(.7,3.8,1,4),line(.7,4.2,1,4)), line(-13,-38,12,37)  )}}}

Edwin</pre>