3y*1/3y=1,y=0

Solution for 3y*1/3y=1,y=0 equation:

3y*1/3y=1.y=0

We move all terms to the left:

3y*1/3y-(1.y)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

3y*1/3y-y=0

We multiply all the terms by the denominator


3y*1-y*3y=0

Wy multiply elements

-3y^2+3y=0

a = -3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-3)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-3}=\frac{-6}{-6}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-3}=\frac{0}{-6}
=0
$