SOLUTION: |(5-2x)/(x+2)| ≤ 1

An inequality

     ≤ 1


is equivalent to

    -1 <=  <= 1.


It means that we should solve TWO inequalities 

    (a)  -1 <=    and   (b)   <= 1

and then take the intersection set of their solutions.


The domain (the set of real numbers where the rational function is defined) is the set of all real numbers except of x = -2.



I will solve the inequality (a)   -1 <=     first.



Let us consider the domain x > -2;  where (x+2) > 0 is positive.


     Then the inequality (a) is equivalent to

        -(x+2) <= 5-2x,   -x - 2 <= 5 - 2x,   2x - x < = 5 +2,   x <= 7.

     So, the interval (-2,7] is the part of the solution to the inequality (a) in the domain x > -2.


We should continue and complete the solution to the inequality (a), now in the domain x < -2.


     Then (x+2) < 0 is negative, and the inequality (a) is equivalent to 

          -(x+2) >= 5-2x,   -x - 2 >= 5-2x,   2x - x >= 5 + 2,   x >= 7.


     So, in the domain x < -2  the inequality (a) has no solution.


Thus we completed solving inequality (a) and found the set of its solutions as the interval (-2,7].



Now we should solve the inequality  (b)    <= 1.


Again let's start with the domain  x > -2.


     In this case, the inequality (b) is equivalent to

        5-2x <= x+2,   5-2 <= x + 2x,   3 < = 3x,   x >= 1.

     So, the interval x >= 1 is the part of the solution to the inequality (b) in the domain x > -2.


We should continue and complete the solution to the inequality (b), now in the domain x < -2.


     Then (x+2) < 0 is negative, and the inequality (b) is equivalent to 

          5-2x >= x+2,   5-2 >= x + 2x,   3 >= 3x,   x <= 1.


     So, the interval x < -2  is the solution to inequality (b) in this case.


Thus we completed solving inequality (b) and found the set of its solutions as the union of intervals (,) U [,).


The intersection of solutions to (a) and (b)  is the intersection of the sets

    (-2,7]   (for (a))   and   (,) U [,)   (for (b)).


This intersection is the set [1,7].


ANSWER.  The solution to the original inequality is the set  [1,7].

    


    Plot y =  (red) and  y = 1 (green)