1/3(2b+9)=2/3(b+9)

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Solution for 1/3(2b+9)=2/3(b+9) equation: 



1/3(2b+9)=2/3(b+9)

We move all terms to the left:

1/3(2b+9)-(2/3(b+9))=0



Domain of the equation: 3(2b+9)!=0

b∈R





Domain of the equation: 3(b+9))!=0

b∈R



We calculate fractions


(3bb/(3(2b+9)*3(b+9)))+(-6b2/(3(2b+9)*3(b+9)))=0



We calculate terms in parentheses: +(3bb/(3(2b+9)*3(b+9))), so:

3bb/(3(2b+9)*3(b+9))

We multiply all the terms by the denominator


3bb

Back to the equation:

+(3bb)





We calculate terms in parentheses: +(-6b2/(3(2b+9)*3(b+9))), so:

-6b2/(3(2b+9)*3(b+9))

We multiply all the terms by the denominator


-6b2

We add all the numbers together, and all the variables

-6b^2

Back to the equation:

+(-6b^2)



We get rid of parentheses

-6b^2+3bb=0

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