-5(y-2y)+7=6(3y-y)+1+16y

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Solution for -5(y-2y)+7=6(3y-y)+1+16y equation: 



-5(y-2y)+7=6(3y-y)+1+16y

We move all terms to the left:

-5(y-2y)+7-(6(3y-y)+1+16y)=0

We add all the numbers together, and all the variables

-5(-1y)-(6(+2y)+1+16y)+7=0

We multiply parentheses

5y-(6(+2y)+1+16y)+7=0



We calculate terms in parentheses: -(6(+2y)+1+16y), so:

6(+2y)+1+16y

determiningTheFunctionDomain
6(+2y)+16y+1

We add all the numbers together, and all the variables

16y+6(+2y)+1

We multiply parentheses

16y+12y+1

We add all the numbers together, and all the variables

28y+1

Back to the equation:

-(28y+1)



We get rid of parentheses

5y-28y-1+7=0

We add all the numbers together, and all the variables

-23y+6=0

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

-23y=-6

y=-6/-23

y=6/23

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