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

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

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

We move all terms to the left:

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


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R


We add all the numbers together, and all the variables

y-(-3y+4)/2y-(+7y)-(+1/5)=0

We get rid of parentheses

y-(-3y+4)/2y-7y-1/5=0

We calculate fractions


y-7y+(15y-20)/10y+(-2y)/10y=0

We add all the numbers together, and all the variables

-6y+(15y-20)/10y+(-2y)/10y=0

We multiply all the terms by the denominator


-6y*10y+(15y-20)+(-2y)=0

Wy multiply elements

-60y^2+(15y-20)+(-2y)=0

We get rid of parentheses

-60y^2+15y-2y-20=0

We add all the numbers together, and all the variables

-60y^2+13y-20=0

a = -60; b = 13; c = -20;
Δ = b2-4ac
Δ = 132-4·(-60)·(-20)
Δ = -4631
Delta is less than zero, so there is no solution for the equation