-5(y+1)+4(3y-4)=6(y-3)+14

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Solution for -5(y+1)+4(3y-4)=6(y-3)+14 equation: 



-5(y+1)+4(3y-4)=6(y-3)+14

We move all terms to the left:

-5(y+1)+4(3y-4)-(6(y-3)+14)=0

We multiply parentheses

-5y+12y-(6(y-3)+14)-5-16=0



We calculate terms in parentheses: -(6(y-3)+14), so:

6(y-3)+14

We multiply parentheses

6y-18+14

We add all the numbers together, and all the variables

6y-4

Back to the equation:

-(6y-4)



We add all the numbers together, and all the variables

7y-(6y-4)-21=0

We get rid of parentheses

7y-6y+4-21=0

We add all the numbers together, and all the variables

y-17=0

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

y=17

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