-3(3y-5)+7y=9+4(4y-2)

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Solution for -3(3y-5)+7y=9+4(4y-2) equation: 



-3(3y-5)+7y=9+4(4y-2)

We move all terms to the left:

-3(3y-5)+7y-(9+4(4y-2))=0

We add all the numbers together, and all the variables

7y-3(3y-5)-(9+4(4y-2))=0

We multiply parentheses

7y-9y-(9+4(4y-2))+15=0



We calculate terms in parentheses: -(9+4(4y-2)), so:

9+4(4y-2)

determiningTheFunctionDomain
4(4y-2)+9

We multiply parentheses

16y-8+9

We add all the numbers together, and all the variables

16y+1

Back to the equation:

-(16y+1)



We add all the numbers together, and all the variables

-2y-(16y+1)+15=0

We get rid of parentheses

-2y-16y-1+15=0

We add all the numbers together, and all the variables

-18y+14=0

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

-18y=-14

y=-14/-18

y=7/9

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