4/3(b-12)+7/3b=7/6

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Solution for 4/3(b-12)+7/3b=7/6 equation: 



4/3(b-12)+7/3b=7/6

We move all terms to the left:

4/3(b-12)+7/3b-(7/6)=0



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

b∈R





Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R



We add all the numbers together, and all the variables

4/3(b-12)+7/3b-(+7/6)=0

We get rid of parentheses

4/3(b-12)+7/3b-7/6=0

We calculate fractions


(126bb/(3(b-12)*3b*6)+(-63b^2b/(3(b-12)*3b*6)+432b/(3(b-12)*3b*6)=0



We calculate terms in parentheses: +(126bb/(3(b-12)*3b*6)+(-63b^2b/(3(b-12)*3b*6)+432b/(3(b-12)*3b*6), so:

126bb/(3(b-12)*3b*6)+(-63b^2b/(3(b-12)*3b*6)+432b/(3(b-12)*3b*6

We can not solve this equation

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