-(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!=0
x!=0/1
x!=0
x∈R
Domain of the equation: 3)x!=0
x!=0/1
x!=0
x∈R
We add all the numbers together, and all the variables
-(+1/6)x+(+1/3)x-(x-2)-12-3=0
We add all the numbers together, and all the variables
-(+1/6)x+(+1/3)x-(x-2)-15=0
We multiply parentheses
-x^2+x^2-(x-2)-15=0
We get rid of parentheses
-x^2+x^2-x+2-15=0
We add all the numbers together, and all the variables
-1x-13=0
We move all terms containing x to the left, all other terms to the right
-x=13
x=13/-1
x=-13

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