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

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



((3x-2)/3)-((2x+1)/6)=x-1
We move all terms to the left:
((3x-2)/3)-((2x+1)/6)-(x-1)=0
We get rid of parentheses
((3x-2)/3)-((2x+1)/6)-x+1=0
We calculate fractions
-x+(18x-12)/()+(-6x-3)/()+1=0
We add all the numbers together, and all the variables
-1x+(18x-12)/()+(-6x-3)/()+1=0
We multiply all the terms by the denominator
-1x*()+(18x-12)+(-6x-3)+1*()=0
We add all the numbers together, and all the variables
-1x*()+(18x-12)+(-6x-3)=0
We get rid of parentheses
-1x*()+18x-6x-12-3=0
We add all the numbers together, and all the variables
12x-1x*()-15=0
We move all terms containing x to the left, all other terms to the right
12x-1x*()=15

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