Question 1162206
<pre>
{{{matrix(1,3,
1/(x-2),"">="",1)}}}

Get 0 on the right:

{{{matrix(1,3,
1/(x-2)-1,"">="",0)}}}

Get LCD

{{{matrix(1,3,
1/(x-2)-(x-2)/(x-2),"">="",0)}}}

{{{matrix(1,3,
(1-(x-2))/(x-2),"">="",0)}}}

{{{matrix(1,3,
(1-x+2)/(x-2),"">="",0)}}}

{{{matrix(1,3,
(3-x)/(x-2),"">="",0)}}}

The critical values are when the numerator or
the denominator equals 0:

The numerator is 0 when x = 3
The denominator is 0 when x = 2

So we draw a number line and mark the critical 
numbers

-----------------o---o- ---------
-2  -1   0   1   2   3   4   5  6

This divides the number line into three intervals

We pick a test value in the left-most interval.
The easiest number there to pick for a test value
is 0, so we substitute it in the inequality:

{{{matrix(1,3,
(3-x)/(x-2),"">="",0)}}}
{{{matrix(1,3,
(3-0)/(0-2),"">="",0)}}}
{{{matrix(1,3,
-3/2,"">="",0)}}}

That's false so we do not shade the numbers in the
left most region, so we still have

-----------------o---o-----------
-2  -1   0   1   2   3   4   5  6

We pick a test value in the middle interval, between
2 and 3. The easiest number there to pick for a test value
is 2.5, so we substitute it in the inequality:

{{{matrix(1,3,
(3-x)/(x-2),"">="",0)}}}
{{{matrix(1,3,
(3-2.5)/(2.5-2),"">="",0)}}}
{{{matrix(1,3,
0.5/0.5,"">="",0)}}}
{{{matrix(1,3,
1,"">="",0)}}}

That's true so we shade the numbers in the
middle region, so we now have

-----------------o===o---------- 
-2  -1   0   1   2   3   4   5  6


We pick a test value in the right-most interval, greater
than 3. The easiest number there to pick for a test value
is 4, so we substitute it in the inequality:

{{{matrix(1,3,
(3-x)/(x-2),"">="",0)}}}
{{{matrix(1,3,
(3-4)/(4-2),"">="",0)}}}
{{{matrix(1,3,
-1/2,"">="",0)}}}
{{{matrix(1,3,
1,"">="",0)}}}

That's false so we do not shade the numbers in the
right-most region, so we have

-----------------o===o----------- 
-2  -1   0   1   2   3   4   5  6

Finally we test the critical numbers themselves:

We test 2

{{{matrix(1,3,
(3-x)/(x-2),"">="",0)}}}
{{{matrix(1,3,
(3-2)/(2-2),"">="",0)}}}
{{{matrix(1,3,
1/0,"">="",1)}}}

1/0 is not defined, so we do not shade at 2

We test 3

{{{matrix(1,3,
(3-x)/(x-2),"">="",0)}}}
{{{matrix(1,3,
(3-3)/(3-2),"">="",0)}}}
{{{matrix(1,3,
0/1,"">="",0)}}}
{{{matrix(1,3,
0,"">="",0)}}}

That's true so we do shade at 3

-----------------o===☻----------- 
-2  -1   0   1   2   3   4   5  6

Interval notation:

{{{matrix(1,5,

"(",2,",",3,"]")}}}

Edwin</pre>