(b+2)/3=9-(b+6)/2

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Solution for (b+2)/3=9-(b+6)/2 equation: 



(b+2)/3=9-(b+6)/2

We move all terms to the left:

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

We calculate fractions


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



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

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

We multiply all the terms by the denominator


-(9-(b+6)*3)



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

9-(b+6)*3

determiningTheFunctionDomain
-(b+6)*3+9

We multiply parentheses

-3b-18+9

We add all the numbers together, and all the variables

-3b-9

Back to the equation:

-(-3b-9)



We get rid of parentheses

3b+9

Back to the equation:

+(3b+9)



We get rid of parentheses

b/()+3b+9=0

We multiply all the terms by the denominator


b+3b*()+9*()=0

We add all the numbers together, and all the variables

b+3b*()=0

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