Question 1137569
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<pre>
{{{(2x)/(x+2)}}} >= {{{(x)/(x-2)}}}

has the domain  x =/= -2  and  x =/= 2  and is equivalent to (moving the right part to the left)

{{{(2x)/(x+2)}}} - {{{(x)/(x-2)}}} >= 0


Write with the common denominator

{{{((2x)*(x-2) - x*(x+2))/((x+2)*(x-2))}}} >= 0.



Simplify step by step

{{{(2x^2 - 2x - x^2 - 2x)/((x+2)*(x-2))}}} >= 0,

{{{(x^2 - 4x)/((x+2)*(x-2))}}} >= 0,

{{{(x*(x-4))/((x+2)*(x-2))}}} >= 0.      (1)



Simplifying is completed. Next we start analyzing.


The critical points, where the rational function (1) changes its sign, are the points  -2, 0, 2 and 4.


They divide the number line in 5 intervals.


1)  Interval  ({{{-infinity}}},{{{-2}}}).  

    All four separate factors are negative, hence, the function is POSITIVE.

    Thus this interval <U>IS</U> the part of the solution.



2)  Interval  ({{{-2}}},{{{0}}}).  

    One factor, (x+2) is positive; the other 3 factors are negative; hence, the function is NEGATIVE.

    Thus this interval is <U>NOT</U> the part of the solution.



3)  Interval  [{{{0}}},{{{2}}}).  

    Two factors,  (x+2) and x,  are positive; the other 2 factors are negative; hence, the function is POSITIVE.

    Thus this interval <U>IS</U> the part of the solution.



Moving forward along the number line interval after interval, and analyzing by the similar way, you get the 


<U>ANSWER</U>.  The solution is the union of the following intervals:


         ({{{-infinity}}},{{{-2}}}) U [{{{0}}},{{{2}}}) U [{{{4}}},{{{infinity}}}).
</pre>

Completed, explained and solved.



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Do you understand now why the forum requires one and only one problem per post ?


Because otherwise you will get a mess.