SOLUTION: How would I solve (3x+2/x+1) > 4 ? Thank you for your help

, with  

3x + 2 > 4(x + 1) ------- Multiplying by LCD, x + 1

3x + 2 > 4x + 4

3x - 4x > 4 - 2

- x > 2

 ------> 



We now have 2 critical points: - 2 and - 1, and 3 INTERVALS to check:

1)   x < - 2, or x = - 3 (- 3 was chosen since - 3 < - 2)

2)   - 2 < x < - 1, or x = - 1.5 (this is a value between - 2 and - 1)

3)   x > - 1, or x = 0 (0 was chosen since 0 > - 1)



1)   x < - 2, or x = - 3

      ----- Substituting - 3 in original inequality

     

      ------- False, so x < - 2 is NOT a solution, as this interval DOES NOT satisfy the inequality



2)   - 2 < x < - 1, or x = - 1.5

      ----- Substituting - 1.5 in original inequality

     

      ------- TRUE, so  IS a solution, as this interval DOES satisfy the inequality

 

3)   x > - 1, or x = 0

      ----- Substituting 0 in original inequality

     

      ------- False, so x > - 1 is NOT a solution, as this interval DOES NOT satisfy the inequality 

1. First, let us assume that x+1 > 0.
   In other words, we will consider now real numbers { x | x > -1 }. 

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

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

   Thus we obtain this: if x > -1, then x < -2. 
   
   It is, surely, absurd. 
   So, in the domain x > -1 there is no solution to (1).


2. Next, let us consider the interval x < -1. In this interval, the denominator (x+1) is negative.

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

   3x+2 < 4*(x+1).     (2) 

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

   Further, (2) implies  3x+2 < 4x+4  --->  2-4 < 4x-3x  --->  -2 < x,   or   x > -2.
   
   Thus we obtain this: if x < -1, then x > -2.

   It means that the set of real numbers -2 < x < -1 satisfies the inequality (1).

   It is the solution of the inequality (1).

Answer. The solution to (1) is the interval (-2,-1).

Figure 1. Plot y =