SOLUTION: How would this be solved: {{{(3x+1)/(x-2) > 2}}} Thank you

The other tutor's method is incorrect.  He multiplied through
by x and assumed that x is positive and only gave one case.  
He forgot that x might be negative.  His method
would require two cases, one when x is positive and the 
inequality remains the same and a second case where x is negative
and the inequality is reversed.  So his method only gave part of
the solution.  Here is the usual way to tackle such an inequality
as this.  I think you meant this rather than the way the other
tutor interpreted your inequality.  You should have placed 
parentheses around numerators and denominators to show where they
begin and end.  Otherwise we go by the PEMDAS rule and that may
not be what you intended.



Get 0 on the right by subtracting 2 from both sides



Get a common denominator











The critical numbers are found by setting
the numerator and the denominators = 0:

x+5=0 gives critical number -5
x-2=0 gives critical number 2

Put those on a number line:

----------o-----------------o---------
-8 -7 -6 -5 -4 -3 -2  0  1  2  3  4  5

We test a number in each of the three intervals:

in the interval x < -5 we choose test value -6
and substitute it in






Since that is true, we shade the leftmost interval

<=========o-----------------o---------
-8 -7 -6 -5 -4 -3 -2  0  1  2  3  4  5

In the interval -5 < x < 2 we choose test value 0
and substitute it in






Since that is false, we do not shade that interval,
so we still have

<=========o-----------------o---------
-8 -7 -6 -5 -4 -3 -2  0  1  2  3  4  5

in the interval x > 2 we choose test value 3
and substitute it in






Since that is true, we shade the rightmostmost interval

<=========o-----------------o========>
-8 -7 -6 -5 -4 -3 -2  0  1  2  3  4  5

Finally we test the critical numbers.

Testing critical number -5






That's false, so we do not darken the circle
at -5

Testing critical number 2






That's false, so we do not darken the circle
at 2 either.

So the graph of the solution is

<=========o-----------------o========>
-8 -7 -6 -5 -4 -3 -2  0  1  2  3  4  5

which in interval notation is

  

or in set builder notation

{x | x < -5 or x > 2 }


Edwin

 
  

  

 > 2.      (1)


1. First, let us assume that x-2 > 0.
   In other words, we will look now for solutions in the domain  { x | x > 2 }. 

   Multiply both side of (1) by (x-2), which is positive in this case. Then you will get inequality           

   3x+1 > 2*(x-2)  --->  3x+1 > 2x-4  --->  3x - 2x > -4 -1  --->  x > -5.

   Thus we obtain this: the solution is the intersection of two sets: {x| x > 2} and {x| x > -5}. 
   
   This intersection is the set {x| x > 2}.
   
   So, the interval (,) is the solution to (1), under the assumption that x > 2.


2. Next, let us consider the domain x < 2. In this domain, the denominator (x-2) is negative.

   Multiply both side of (1) by (x-2), which is negative now. Then you will get

   3x+1 < 2*(x-2).     (2) 

   Notice, that I changed the sign ">" of the original inequality to the opposite sign "<", when 
   multiplied both sides of (1) by negative number (x-2).

   Now, (2) implies  3x+1 < 2x-4  --->  3x - 2x < -4 -1  --->  x < -5.
   
   Thus by analyzing the domain x < 2 we obtain the solution  x < -5.

   By collecting the results of n.1 and n.2 you get the full solution set.
   It is the union (,) U (,).

   The problem is solved.

Answer. The solution to  (1)  is the union  (,) U (,).


The plot of the function    is shown in Figure.

 
  





Figure. Plot y =