(2z+7)/5=(9-3z)/6

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Solution for (2z+7)/5=(9-3z)/6 equation: 



(2z+7)/5=(9-3z)/6

We move all terms to the left:

(2z+7)/5-((9-3z)/6)=0

We add all the numbers together, and all the variables

(2z+7)/5-((-3z+9)/6)=0

We calculate fractions


2z/()+(-((-3z+9)*5)/()=0



We calculate terms in parentheses: +(-((-3z+9)*5)/(), so:

-((-3z+9)*5)/(

We multiply all the terms by the denominator


-((-3z+9)*5)



We calculate terms in parentheses: -((-3z+9)*5), so:

(-3z+9)*5

We multiply parentheses

-15z+45

Back to the equation:

-(-15z+45)



We get rid of parentheses

15z-45

Back to the equation:

+(15z-45)



We get rid of parentheses

2z/()+15z-45=0

We multiply all the terms by the denominator


2z+15z*()-45*()=0

We add all the numbers together, and all the variables

2z+15z*()=0

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