3/4(20y-8)+5=1/2y+20y+8)

Solution for 3/4(20y-8)+5=1/2y+20y+8) equation:

3/4(20y-8)+5=1/2y+20y+8)

We move all terms to the left:

3/4(20y-8)+5-(1/2y+20y+8))=0


Domain of the equation: 4(20y-8)!=0

y∈R



Domain of the equation: 2y+20y+8))+5!=0

y∈R


We add all the numbers together, and all the variables

3/4(20y-8)-(20y+1/2y+8))+5=0

We calculate fractions


6y/(160y^2-64y)+(-(20y+4y2)/(160y^2-64y)=0


We calculate terms in parentheses: +(-(20y+4y2)/(160y^2-64y), so:

-(20y+4y2)/(160y^2-64y

We add all the numbers together, and all the variables

-(+20y+4y^2)/(160y^2-64y

We multiply all the terms by the denominator


-(+20y+4y^2)

We get rid of parentheses

-4y^2-20y

Back to the equation:

+(-4y^2-20y)


We get rid of parentheses

-4y^2-20y+6y/(160y^2-64y)=0

We multiply all the terms by the denominator


-4y^2*(160y^2-64y)-20y*(160y^2-64y)+6y=0

We add all the numbers together, and all the variables

-4y^2*(160y^2-64y)+6y-20y*(160y^2-64y)=0

We multiply parentheses

-4y^2*(160y^2-64y)-3200y^3+1280y^2+6y=0

We do not support eypression: y^3