-3+y=1/3y

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

-3+y=1/3y

We move all terms to the left:

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


Domain of the equation: 3y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

y-1/3y-3=0

We multiply all the terms by the denominator


y*3y-3*3y-1=0

Wy multiply elements

3y^2-9y-1=0

a = 3; b = -9; c = -1;
Δ = b2-4ac
Δ = -92-4·3·(-1)
Δ = 93
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{93}}{2*3}=\frac{9-\sqrt{93}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{93}}{2*3}=\frac{9+\sqrt{93}}{6}
$