3/7(x-8)=1/6(2x+4)

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Solution for 3/7(x-8)=1/6(2x+4) equation: 



3/7(x-8)=1/6(2x+4)

We move all terms to the left:

3/7(x-8)-(1/6(2x+4))=0



Domain of the equation: 7(x-8)!=0

x∈R





Domain of the equation: 6(2x+4))!=0

x∈R



We calculate fractions


(18x2/(7(x-8)*6(2x+4)))+(-7xx/(7(x-8)*6(2x+4)))=0



We calculate terms in parentheses: +(18x2/(7(x-8)*6(2x+4))), so:

18x2/(7(x-8)*6(2x+4))

We multiply all the terms by the denominator


18x2

We add all the numbers together, and all the variables

18x^2

Back to the equation:

+(18x^2)





We calculate terms in parentheses: +(-7xx/(7(x-8)*6(2x+4))), so:

-7xx/(7(x-8)*6(2x+4))

We multiply all the terms by the denominator


-7xx

Back to the equation:

+(-7xx)



We get rid of parentheses

18x^2-7xx=0

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