(7/3x)+(1/3x)=1+(5/3x)

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Solution for (7/3x)+(1/3x)=1+(5/3x) equation: 



(7/3x)+(1/3x)=1+(5/3x)

We move all terms to the left:

(7/3x)+(1/3x)-(1+(5/3x))=0



Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R





Domain of the equation: 3x))!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

(+7/3x)+(+1/3x)-(1+(+5/3x))=0

We get rid of parentheses

7/3x+1/3x-(1+(+5/3x))=0

We calculate fractions


3x/9x^2+(-(1+(+5*3x)/9x^2=0

We multiply all the terms by the denominator


3x+(-(1+(+5*3x)=0



We calculate terms in parentheses: +(-(1+(+5*3x), so:

-(1+(+5*3x





We calculate terms in parentheses: +(-(1+(+5*3x), so:

-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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-(1+(+5*3x





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