1/6(x-12)+1/3(x+3)=x-2

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Solution for 1/6(x-12)+1/3(x+3)=x-2 equation: 



1/6(x-12)+1/3(x+3)=x-2

We move all terms to the left:

1/6(x-12)+1/3(x+3)-(x-2)=0



Domain of the equation: 6(x-12)!=0

x∈R





Domain of the equation: 3(x+3)!=0

x∈R



We get rid of parentheses

1/6(x-12)+1/3(x+3)-x+2=0

We calculate fractions


-x+(3xx/(6(x-12)*3(x+3))+(6xx/(6(x-12)*3(x+3))+2=0

We can not solve this equation

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