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