6(y-4)-3(5y+1)=4(y-2)-4(2y+1)

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Solution for 6(y-4)-3(5y+1)=4(y-2)-4(2y+1) equation: 



6(y-4)-3(5y+1)=4(y-2)-4(2y+1)

We move all terms to the left:

6(y-4)-3(5y+1)-(4(y-2)-4(2y+1))=0

We multiply parentheses

6y-15y-(4(y-2)-4(2y+1))-24-3=0



We calculate terms in parentheses: -(4(y-2)-4(2y+1)), so:

4(y-2)-4(2y+1)

We multiply parentheses

4y-8y-8-4

We add all the numbers together, and all the variables

-4y-12

Back to the equation:

-(-4y-12)



We add all the numbers together, and all the variables

-9y-(-4y-12)-27=0

We get rid of parentheses

-9y+4y+12-27=0

We add all the numbers together, and all the variables

-5y-15=0

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

-5y=15

y=15/-5

y=-3

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