y-1/y+1=y+1/3y-1

Solution for y-1/y+1=y+1/3y-1 equation:

y-1/y+1=y+1/3y-1

We move all terms to the left:

y-1/y+1-(y+1/3y-1)=0


Domain of the equation: y!=0

y∈R



Domain of the equation: 3y-1)!=0

y∈R


We get rid of parentheses

y-1/y-y-1/3y+1+1=0

We calculate fractions


y-y+(-3y)/3y^2+(-y)/3y^2+1+1=0

We add all the numbers together, and all the variables

y-y+(-3y)/3y^2+(-1y)/3y^2+1+1=0

We add all the numbers together, and all the variables

(-3y)/3y^2+(-1y)/3y^2+2=0

We multiply all the terms by the denominator


(-3y)+(-1y)+2*3y^2=0

Wy multiply elements

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

We get rid of parentheses

6y^2-3y-1y=0

We add all the numbers together, and all the variables

6y^2-4y=0

a = 6; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·6·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*6}=\frac{0}{12}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*6}=\frac{8}{12}
=2/3
$