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Q. The solution set of the in equation $ \frac{x+11}{x-3}>0 $ is

KEAMKEAM 2010Linear Inequalities

Solution:

$ \frac{x+11}{x-3}>0 $
$ \Rightarrow $ $ (x+11)(x-3)>0 $
$ \Rightarrow $ $ x+11>0 $ and $ (x-3)>0 $ or $ x+11<0 $ and $ x-3<0 $
$ \Rightarrow $ $ x>-11 $ and $ x>3 $ or $ x<-11 $ and $ x<3 $
$ \Rightarrow $ $ x<-11 $ and $ x>3 $

$ \Rightarrow $ $ x\in (-\infty ,-11)\cup (3,\infty ) $

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