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1/3(x-2)=1/5(x+4)
We move all terms to the left:
1/3(x-2)-(1/5(x+4))=0
Domain of the equation: 3(x-2)!=0
x∈R
Domain of the equation: 5(x+4))!=0We calculate fractions
x∈R
(5xx/(3(x-2)*5(x+4)))+(-3xx/(3(x-2)*5(x+4)))=0
We calculate terms in parentheses: +(5xx/(3(x-2)*5(x+4))), so:
5xx/(3(x-2)*5(x+4))
We multiply all the terms by the denominator
5xx
Back to the equation:
+(5xx)
We calculate terms in parentheses: +(-3xx/(3(x-2)*5(x+4))), so:We get rid of parentheses
-3xx/(3(x-2)*5(x+4))
We multiply all the terms by the denominator
-3xx
Back to the equation:
+(-3xx)
5xx-3xx=0
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