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

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

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

We move all terms to the left:

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


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: (3+3y)!=0

We move all terms containing y to the left, all other terms to the right

3y!=-3

y!=-3/3

y!=-1

y∈R


We add all the numbers together, and all the variables

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

We calculate fractions


(6y+6)/(9y^2+9y)+(-3y)/(9y^2+9y)-2=0

We multiply all the terms by the denominator


(6y+6)+(-3y)-2*(9y^2+9y)=0

We multiply parentheses

-18y^2+(6y+6)+(-3y)-18y=0

We get rid of parentheses

-18y^2+6y-3y-18y+6=0

We add all the numbers together, and all the variables

-18y^2-15y+6=0

a = -18; b = -15; c = +6;
Δ = b2-4ac
Δ = -152-4·(-18)·6
Δ = 657
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{73}}{2*-18}=\frac{15-3\sqrt{73}}{-36}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{73}}{2*-18}=\frac{15+3\sqrt{73}}{-36}
$