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 Dec 8, 2018
 #1
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The first choice and fourth choice. 

 

We could find the least common multiplier, turn things around and get \(x\left(x+3\right)+2=5\left(x+2\right)\) .

 

Then, solving, we get \(x=4,\:x=-2\) and \(\boxed{x=4}\)  as    \(x=2\) , creates a zero in the denominator. 

 Dec 8, 2018

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