5(y-3)=7y+1-2(-6y-1)

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Solution for 5(y-3)=7y+1-2(-6y-1) equation: 



5(y-3)=7y+1-2(-6y-1)

We move all terms to the left:

5(y-3)-(7y+1-2(-6y-1))=0

We multiply parentheses

5y-(7y+1-2(-6y-1))-15=0



We calculate terms in parentheses: -(7y+1-2(-6y-1)), so:

7y+1-2(-6y-1)

determiningTheFunctionDomain
7y-2(-6y-1)+1

We multiply parentheses

7y+12y+2+1

We add all the numbers together, and all the variables

19y+3

Back to the equation:

-(19y+3)



We get rid of parentheses

5y-19y-3-15=0

We add all the numbers together, and all the variables

-14y-18=0

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

-14y=18

y=18/-14

y=-1+2/7

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