3(y+1)+4=2(y+1)+y+y

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



3(y+1)+4=2(y+1)+y+y

We move all terms to the left:

3(y+1)+4-(2(y+1)+y+y)=0

We multiply parentheses

3y-(2(y+1)+y+y)+3+4=0



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

2(y+1)+y+y

We add all the numbers together, and all the variables

2y+2(y+1)

We multiply parentheses

2y+2y+2

We add all the numbers together, and all the variables

4y+2

Back to the equation:

-(4y+2)



We add all the numbers together, and all the variables

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

We get rid of parentheses

3y-4y-2+7=0

We add all the numbers together, and all the variables

-1y+5=0

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

-y=-5

y=-5/-1

y=+5

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