Question 524532
<pre>

You probably did this:


   -3 = -2x + 1

   2x = 4

    x = 2

Now pretend that equation has a term "0y", 
so you can write it this way:

   x + 0y = 2

Now make a table of values:

     x | y
       |
       |
       |

But only substitute arbitrary values for y and solve for x:

Let y = 4, -3, 1, and 7

     x | y
       | 4
       |-3
       | 1
       | 7

 Substitute y=4 in  x + 0y = 2
                  x + 0(4) = 2
                     x + 0 = 2
                         x = 2

Substitute y=-3 in  x + 0y = 2
                 x + 0(-3) = 2
                     x + 0 = 2
                         x = 2

 Substitute y=1 in  x + 0y = 2
                  x + 0(1) = 2
                     x + 0 = 2
                         x = 2

 Substitute y=7 in  x + 0y = 2
                  x + 0(7) = 2
                     x + 0 = 2
                         x = 2

So we have this table of values:

     x | y
     2 | 4
     2 |-3
     2 | 1
     2 | 7

So we plot the points (2,4), (2,-3), (2,1) and (2,7)

{{{drawing(400,400,-8,8,-8,8, graph(400,400,-8,8,-8,8),
locate(2,4,"(2,4)"),locate(2,-3,"(2,-3)"),locate(2,1,"(2,1)"),locate(2,7,"(2,7)"),

circle(2,4,.1), circle(2,-3,.1), circle(2,1,.1), circle(2,7,.1) )}}}

Then connect the points with a straight line:

{{{drawing(400,400,-8,8,-8,8, graph(400,400,-8,8,-8,8),
locate(2,4,"(2,4)"),locate(2,-3,"(2,-3)"),locate(2,1,"(2,1)"),locate(2,7,"(2,7)"),line(2,-10,2,10),

circle(2,4,.1), circle(2,-3,.1), circle(2,1,.1), circle(2,7,.1) )}}}
    
So you see, it's just a vertical line.  When the equation does not
contain a y, and is of the form

x = b

the graph is a vertical line that goes through (b,0).  You don't need
to plot a bunch of points.  I just did it to show you why it works.

Edwin</pre>