(2z-4)/6=(2z+7)/5

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Solution for (2z-4)/6=(2z+7)/5 equation: 



(2z-4)/6=(2z+7)/5

We move all terms to the left:

(2z-4)/6-((2z+7)/5)=0

We calculate fractions


2z/()+(-((2z+7)*6)/()=0



We calculate terms in parentheses: +(-((2z+7)*6)/(), so:

-((2z+7)*6)/(

We multiply all the terms by the denominator


-((2z+7)*6)



We calculate terms in parentheses: -((2z+7)*6), so:

(2z+7)*6

We multiply parentheses

12z+42

Back to the equation:

-(12z+42)



We get rid of parentheses

-12z-42

Back to the equation:

+(-12z-42)



We get rid of parentheses

2z/()-12z-42=0

We multiply all the terms by the denominator


2z-12z*()-42*()=0

We add all the numbers together, and all the variables

2z-12z*()=0

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