Question 26497
(x-2)(x+3)/(x-4) > 0
FOR A FRACTION TO BE POSITIVE ,BOTH N.R AND D.R SHOULD BE POSTIVE OR NEGATIVE
CASE I....N.R AND D.R POSITIVE
FOR PRODUCT OF 2 FACTORS TO BE POSITIVE BOTH SHOUD BE POSITIVE OR NEGATIVE.
VARIANT I
X-2,X+3 ARE POSITIVE.SO X>2..D.R=X-4....X-4>0...SO X>4
THAT IS IN THIS CASE X>4
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VARIANT II
X-2,X+3 ARE NEGATIVE.SO X<-3....D.R=X-4.....X-4>0...SO X>4...THIS IS NOT POSSIBLE
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SO UNDER CASE I,THE ONLY POSSIBILITY IS X>4
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CASE II........N.R AND D.R NEGATIVE.........
FOR PRODUCT OF 2 FACTORS TO BE NEGATIVE ONE SHOUD BE POSITIVE AND ANOTHER NEGATIVE.
SO X SHOULD BE BETWEEN 2 AND -3....AND D.R=X-4<0...X<4
SO X IS BETWEEN 2 AND -3

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hence the total solution set is for 
x to be greater than 4 or between 2 and -3.