Question 949423
<pre>
a. -3&#8804; 3-5x/3 <3/8
b. 3x-5/x >0

When typing algebra, you must put parentheses around 
numerators and denominators if they contain more than 
1 letter or 1 number, like this

a. -3&#8804; (3-5x)/3 <3/8
b. (3x-5)/x >0

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

{{{-3<= (3-5x)/3 <3/8}}} 

Multiply thru all 3 sides by 24

{{{-72<= 8(3-5x) <9}}}

{{{-72<=24-40x <9}}}

Add -24 to all 3 sides:

{{{-72<-40x <9}}}

{{{-96<=-40x<-15}}}

Divide through by -40 which reverses
the inequalities

{{{(-96)/(-40)>=x>(-15)/(-40)}}}

{{{12/5>=x>-3/8}}}

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

b. (3x-5)/x >0

Critical values are when numerator equas zero and when
denominator equals zero.

Numerator=0
3x-5 = 0
  3x = 5
   x = {{{5/3}}} = {{{1&2/3}}}   <-- critical value

Denominator = 0
   x = 0

Put critical values on number line:

------------o------o---------------
            0     {{{5/3}}}

Test any value less than 0, say -1, by substituting
in the original:

{{{(3x-5)/x > 0}}}

{{{(3^""(-1)-5)/(-1) > 0}}}

{{{(-3-5)/(-1) > 0}}}

{{{(-8)/(-1)>0}}}

{{{8>0}}}

That's true so we shade line left of 0

<===========o------o---------------
            0      {{{5/3}}}

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

Next test any value between 0 and {{{5/3}}}, say 1, by substituting
in the original:

{{{(3x-5)/x > 0}}}

{{{(3^""(1)-5)/(1) > 0}}}

{{{(3-5) > 0}}}

{{{-2>0}}}


That's false so we do not shade line between 0 and {{{5/3}}}. We
still have

<===========o------o---------------
            0      {{{5/3}}}

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

Test any value right of {{{5/3}}}, say 2, by substituting
in the original:

{{{(3x-5)/x > 0}}}

{{{(3^""(2)-5)/(2) > 0}}}

{{{(6-5)/2 > 0}}}

{{{(1)/2>0}}}

{{{1/2>0}}}

That's true so we shade line right of {{{5/3}}}

S

<===========o------o==========>
            0      {{{5/3}}}

Interval notation 

{{{matrix(1,3,

(matrix(1,3,-infinity,",",0)), U, (matrix(1,3,5/3,",",infinity)) )}}}

Edwin</pre>