3(y+1)+4=(y-1)+y+y

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



3(y+1)+4=(y-1)+y+y

We move all terms to the left:

3(y+1)+4-((y-1)+y+y)=0

We multiply parentheses

3y-((y-1)+y+y)+3+4=0



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

(y-1)+y+y

We add all the numbers together, and all the variables

2y+(y-1)

We get rid of parentheses

2y+y-1

We add all the numbers together, and all the variables

3y-1

Back to the equation:

-(3y-1)



We add all the numbers together, and all the variables

3y-(3y-1)+7=0

We get rid of parentheses

3y-3y+1+7=0

We add all the numbers together, and all the variables

8!=0

There is no solution for this equation

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