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1/3(x+9)=2/3(2x-6)
We move all terms to the left:
1/3(x+9)-(2/3(2x-6))=0
Domain of the equation: 3(x+9)!=0
x∈R
Domain of the equation: 3(2x-6))!=0We calculate fractions
x∈R
(3x2/(3(x+9)*3(2x-6)))+(-6xx/(3(x+9)*3(2x-6)))=0
We calculate terms in parentheses: +(3x2/(3(x+9)*3(2x-6))), so:
3x2/(3(x+9)*3(2x-6))
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: +(-6xx/(3(x+9)*3(2x-6))), so:We get rid of parentheses
-6xx/(3(x+9)*3(2x-6))
We multiply all the terms by the denominator
-6xx
Back to the equation:
+(-6xx)
3x^2-6xx=0
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