z/13=2(3z+1)/9

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Solution for z/13=2(3z+1)/9 equation: 



z/13=2(3z+1)/9

We move all terms to the left:

z/13-(2(3z+1)/9)=0

We calculate fractions


9z^2/()+(-(2(3z+1)*13)/()=0



We calculate terms in parentheses: +(-(2(3z+1)*13)/(), so:

-(2(3z+1)*13)/(

We multiply all the terms by the denominator


-(2(3z+1)*13)



We calculate terms in parentheses: -(2(3z+1)*13), so:

2(3z+1)*13

We multiply parentheses

78z+26

Back to the equation:

-(78z+26)



We get rid of parentheses

-78z-26

Back to the equation:

+(-78z-26)



We get rid of parentheses

9z^2/()-78z-26=0

We multiply all the terms by the denominator


9z^2-78z*()-26*()=0

We add all the numbers together, and all the variables

9z^2-78z*()=0

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