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

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

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

We move all terms to the left:

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


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

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

-y)!=-3

y!=-3/1

y!=-3

y∈R



Domain of the equation: y+2))!=0

y∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

Wy multiply elements

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

We get rid of parentheses

-2y^2+3y=0

a = -2; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-2)·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*-2}=\frac{-6}{-4}
=1+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-2}=\frac{0}{-4}
=0
$