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1/3(x+3)+1/12(x-12=x+7
We move all terms to the left:
1/3(x+3)+1/12(x-12-(x+7)=0
Domain of the equation: 3(x+3)!=0
x∈R
Domain of the equation: 12(x-12-(x+7)!=0We calculate fractions
x∈R
(12xx/(3(x+3)*12(x-12-(x+7))+(3xx/(3(x+3)*12(x-12-(x+7))=0
We calculate terms in parentheses: +(12xx/(3(x+3)*12(x-12-(x+7))+(3xx/(3(x+3)*12(x-12-(x+7)), so:
12xx/(3(x+3)*12(x-12-(x+7))+(3xx/(3(x+3)*12(x-12-(x+7)
We can not solve this equation
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