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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(y(-1y+2)-3y), so:

y(-1y+2)-3y

We add all the numbers together, and all the variables

-3y+y(-1y+2)

We multiply parentheses

-1y^2-3y+2y

We add all the numbers together, and all the variables

-1y^2-1y

Back to the equation:

-(-1y^2-1y)


We add all the numbers together, and all the variables

-1y^2-(-1y^2-1y)+9y-15=0

We get rid of parentheses

-1y^2+1y^2+1y+9y-15=0

We add all the numbers together, and all the variables

10y-15=0

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

10y=15

y=15/10

y=1+1/2