1/3(7b-4)=5b+12

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Solution for 1/3(7b-4)=5b+12 equation: 



1/3(7b-4)=5b+12

We move all terms to the left:

1/3(7b-4)-(5b+12)=0



Domain of the equation: 3(7b-4)!=0

b∈R



We get rid of parentheses

1/3(7b-4)-5b-12=0

We multiply all the terms by the denominator


-5b*3(7b-4)-12*3(7b-4)+1=0

Wy multiply elements

-15b^2(7-36b(7+1=0

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