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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

y-4(y+2)

We multiply parentheses

y-4y-8

We add all the numbers together, and all the variables

-3y-8

Back to the equation:

-(-3y-8)


We add all the numbers together, and all the variables

-3y-(-3y-8)+11=0

We get rid of parentheses

-3y+3y+8+11=0

We add all the numbers together, and all the variables

19!=0

There is no solution for this equation