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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We calculate fractions


y-4y+(15y-20)/10y+(-2y)/10y-3=0

We add all the numbers together, and all the variables

-3y+(15y-20)/10y+(-2y)/10y-3=0

We multiply all the terms by the denominator


-3y*10y+(15y-20)+(-2y)-3*10y=0

Wy multiply elements

-30y^2+(15y-20)+(-2y)-30y=0

We get rid of parentheses

-30y^2+15y-2y-30y-20=0

We add all the numbers together, and all the variables

-30y^2-17y-20=0

a = -30; b = -17; c = -20;
Δ = b2-4ac
Δ = -172-4·(-30)·(-20)
Δ = -2111
Delta is less than zero, so there is no solution for the equation