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1/3(x+5)=2/3(x-9)
We move all terms to the left:
1/3(x+5)-(2/3(x-9))=0
Domain of the equation: 3(x+5)!=0
x∈R
Domain of the equation: 3(x-9))!=0We calculate fractions
x∈R
(3xx/(3(x+5)*3(x-9)))+(-6xx/(3(x+5)*3(x-9)))=0
We calculate terms in parentheses: +(3xx/(3(x+5)*3(x-9))), so:
3xx/(3(x+5)*3(x-9))
We multiply all the terms by the denominator
3xx
Back to the equation:
+(3xx)
We calculate terms in parentheses: +(-6xx/(3(x+5)*3(x-9))), so:We get rid of parentheses
-6xx/(3(x+5)*3(x-9))
We multiply all the terms by the denominator
-6xx
Back to the equation:
+(-6xx)
3xx-6xx=0
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