(1/3)(7b+2)=b-(2/5)

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Solution for (1/3)(7b+2)=b-(2/5) equation:



(1/3)(7b+2)=b-(2/5)
We move all terms to the left:
(1/3)(7b+2)-(b-(2/5))=0
Domain of the equation: 3)(7b+2)!=0
b∈R
We add all the numbers together, and all the variables
(+1/3)(7b+2)-(b-(+2/5))=0
We multiply parentheses ..
(+7b^2+1/3*2)-(b-(+2/5))=0
We calculate fractions
7b^2/()+(-(b-12)/()=0
We calculate terms in parentheses: +(-(b-12)/(), so:
-(b-12)/(
We multiply all the terms by the denominator
-(b-12)
We get rid of parentheses
-b+12
We add all the numbers together, and all the variables
-1b+12
Back to the equation:
+(-1b+12)
We get rid of parentheses
7b^2/()-1b+12=0
We multiply all the terms by the denominator
7b^2-1b*()+12*()=0
We add all the numbers together, and all the variables
7b^2-1b*()=0

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