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

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

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

We move all terms to the left:

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


Domain of the equation: 3(b+12)!=0

b∈R



Domain of the equation: 6(2b-6))!=0

b∈R


We calculate fractions


(12b2/(3(b+12)*6(2b-6)))+(-3bb/(3(b+12)*6(2b-6)))=0


We calculate terms in parentheses: +(12b2/(3(b+12)*6(2b-6))), so:

12b2/(3(b+12)*6(2b-6))

We multiply all the terms by the denominator


12b2

We add all the numbers together, and all the variables

12b^2

Back to the equation:

+(12b^2)



We calculate terms in parentheses: +(-3bb/(3(b+12)*6(2b-6))), so:

-3bb/(3(b+12)*6(2b-6))

We multiply all the terms by the denominator


-3bb

Back to the equation:

+(-3bb)


We get rid of parentheses

12b^2-3bb=0