-(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 add all the numbers together, and all the variables
-(+1/6)(x-12)+(+1/3)(x+3)-(x-2)=0
We get rid of parentheses
-(+1/6)(x-12)+(+1/3)(x+3)-x+2=0
We multiply parentheses ..
-(+x^2+1/6*-12)+(+1/3)(x+3)-x+2=0
We calculate fractions
There is no solution for this equation

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