(Y+7)/3=1+(3y-2)/5

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Solution for (Y+7)/3=1+(3y-2)/5 equation: 



(+7)/3=1+(3Y-2)/5

We move all terms to the left:

(+7)/3-(1+(3Y-2)/5)=0

We add all the numbers together, and all the variables

-(1+(3Y-2)/5)+7/3=0

We calculate fractions


(-(1+(3Y-2)*3)/()+()/()=0



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

-(1+(3Y-2)*3)/()+()/(

We add all the numbers together, and all the variables

-(1+(3Y-2)*3)/()+1

We multiply all the terms by the denominator


-(1+(3Y-2)*3)+1*()



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

1+(3Y-2)*3

determiningTheFunctionDomain
(3Y-2)*3+1

We multiply parentheses

9Y-6+1

We add all the numbers together, and all the variables

9Y-5

Back to the equation:

-(9Y-5)



We add all the numbers together, and all the variables

-(9Y-5)

We get rid of parentheses

-9Y+5

Back to the equation:

+(-9Y+5)



We get rid of parentheses

-9Y+5=0

We move all terms containing Y to the left, all other terms to the right

-9Y=-5

Y=-5/-9

Y=5/9

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