(5/7z)-(3/7z)+(9/7z)=(z/7)+4

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Solution for (5/7z)-(3/7z)+(9/7z)=(z/7)+4 equation: 



(5/7z)-(3/7z)+(9/7z)=(z/7)+4

We move all terms to the left:

(5/7z)-(3/7z)+(9/7z)-((z/7)+4)=0



Domain of the equation: 7z)!=0

z!=0/1

z!=0

z∈R



We add all the numbers together, and all the variables

(+5/7z)-(+3/7z)+(+9/7z)-((+z/7)+4)=0

We get rid of parentheses

5/7z-3/7z+9/7z-((+z/7)+4)=0

We calculate fractions


()/49z^2+(-((+z*7z)/49z^2=0

We multiply all the terms by the denominator


(-((+z*7z)+()=0



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