Question 751237
<pre>
{{{(x+3)/2x}}} < 2

Subtract 2 from both sides to get 0 on the right:

{{{(x+3)/2x}}} - 2 < 0

Get a common denominator of 2x.  Multiply the 2 by {{{2x/(2x)}}}

{{{(x+3)/(2x)}}} - 2·{{{2x/(2x)}}} < 0

{{{(x+3)/(2x)}}} - {{{4x/(2x)}}} < 0

{{{((x+3)-4x)/(2x)}}} < 0

{{{(x+3-4x)/(2x)}}} < 0

{{{(3-3x)/(2x)}}} < 0

Factor the numerator

{{{(3(1-x))/(2x)}}} < 0

Divide both sides by 3

{{{(1-x)/(2x)}}} < 0

Multiply both sides by 2

{{{(1-x)/x}}} < 0

Find all critical values by setting numerator and denominator = 0

 1-x=0        x=0
  -x=-1       
   x=1  

So the critical values are 0 and 1

Put these on a number line:

----------o--o---------
-3 -2 -1  0  1  2  3  4

test a value to the left of 0, say x=-1

Substitute it in

{{{(1-x)/x}}} < 0

{{{(1-(-1))/(-1)}}} < 0

{{{(1+1)/(-1)}}} < 0

{{{2/(-1)}}} < 0

-2 < 0

This is true so we shade the portion of the
number line left of 0

< ==========o--o---------
  -3 -2 -1  0  1  2  3  4

test a value between 0 and 1 say x=0.5

Substitute it in

{{{(1-0.5)/0.5}}} < 0

{{{0.5/0.5}}} < 0

1 < 0

That is false so we do not shade the portion of the
number line between 0 and 1.  So we still have

< ==========o--o---------
  -3 -2 -1  0  1  2  3  4

We test a value to the right of 1, say x=2

Substitute it in

{{{(1-x)/x}}} < 0

{{{(1-(2))/2}}} < 0

{{{(-1)/2}}} < 0

{{{-1/2}}} < 0


This is true so we shade the portion of the
number line right of 1

<  =========o--o======== >
  -3 -2 -1  0  1  2  3  4

That's the graph of the solution set, in interval notation
that is

{{{(matrix(1,3,

-infinity,",",0))}}}{{{U}}}{{{(matrix(1,3,

1,",",infinity))}}}

Edwin</pre>