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