-3(2y-2)+8y=7+3(4y-1)

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Solution for -3(2y-2)+8y=7+3(4y-1) equation:



-3(2y-2)+8y=7+3(4y-1)
We move all terms to the left:
-3(2y-2)+8y-(7+3(4y-1))=0
We add all the numbers together, and all the variables
8y-3(2y-2)-(7+3(4y-1))=0
We multiply parentheses
8y-6y-(7+3(4y-1))+6=0
We calculate terms in parentheses: -(7+3(4y-1)), so:
7+3(4y-1)
determiningTheFunctionDomain 3(4y-1)+7
We multiply parentheses
12y-3+7
We add all the numbers together, and all the variables
12y+4
Back to the equation:
-(12y+4)
We add all the numbers together, and all the variables
2y-(12y+4)+6=0
We get rid of parentheses
2y-12y-4+6=0
We add all the numbers together, and all the variables
-10y+2=0
We move all terms containing y to the left, all other terms to the right
-10y=-2
y=-2/-10
y=1/5

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