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

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Solution for 5(y-3)+1=3y+4(y+4) equation: 



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

We move all terms to the left:

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

We multiply parentheses

5y-(3y+4(y+4))-15+1=0



We calculate terms in parentheses: -(3y+4(y+4)), so:

3y+4(y+4)

We multiply parentheses

3y+4y+16

We add all the numbers together, and all the variables

7y+16

Back to the equation:

-(7y+16)



We add all the numbers together, and all the variables

5y-(7y+16)-14=0

We get rid of parentheses

5y-7y-16-14=0

We add all the numbers together, and all the variables

-2y-30=0

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

-2y=30

y=30/-2

y=-15

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