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1/2(x+1)=1/3(2x-3)
We move all terms to the left:
1/2(x+1)-(1/3(2x-3))=0
Domain of the equation: 2(x+1)!=0
x∈R
Domain of the equation: 3(2x-3))!=0We calculate fractions
x∈R
(3x2/(2(x+1)*3(2x-3)))+(-2xx/(2(x+1)*3(2x-3)))=0
We calculate terms in parentheses: +(3x2/(2(x+1)*3(2x-3))), so:
3x2/(2(x+1)*3(2x-3))
We multiply all the terms by the denominator
3x2
We add all the numbers together, and all the variables
3x^2
Back to the equation:
+(3x^2)
We calculate terms in parentheses: +(-2xx/(2(x+1)*3(2x-3))), so:We get rid of parentheses
-2xx/(2(x+1)*3(2x-3))
We multiply all the terms by the denominator
-2xx
Back to the equation:
+(-2xx)
3x^2-2xx=0
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