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5/7(x+1)=1/28(x+2)
We move all terms to the left:
5/7(x+1)-(1/28(x+2))=0
Domain of the equation: 7(x+1)!=0
x∈R
Domain of the equation: 28(x+2))!=0We calculate fractions
x∈R
(140xx/(7(x+1)*28(x+2)))+(-7xx/(7(x+1)*28(x+2)))=0
We calculate terms in parentheses: +(140xx/(7(x+1)*28(x+2))), so:
140xx/(7(x+1)*28(x+2))
We multiply all the terms by the denominator
140xx
Back to the equation:
+(140xx)
We calculate terms in parentheses: +(-7xx/(7(x+1)*28(x+2))), so:We get rid of parentheses
-7xx/(7(x+1)*28(x+2))
We multiply all the terms by the denominator
-7xx
Back to the equation:
+(-7xx)
140xx-7xx=0
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