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