3y+2(y-5)=(-3y+35)-y

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Solution for 3y+2(y-5)=(-3y+35)-y equation: 



3y+2(y-5)=(-3y+35)-y

We move all terms to the left:

3y+2(y-5)-((-3y+35)-y)=0

We multiply parentheses

3y+2y-((-3y+35)-y)-10=0



We calculate terms in parentheses: -((-3y+35)-y), so:

(-3y+35)-y

We add all the numbers together, and all the variables

-1y+(-3y+35)

We get rid of parentheses

-1y-3y+35

We add all the numbers together, and all the variables

-4y+35

Back to the equation:

-(-4y+35)



We add all the numbers together, and all the variables

5y-(-4y+35)-10=0

We get rid of parentheses

5y+4y-35-10=0

We add all the numbers together, and all the variables

9y-45=0

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

9y=45

y=45/9

y=5

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