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

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

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

We move all terms to the left:

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


Domain of the equation: y+1)!=0

y∈R



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

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

y!=-2

y∈R



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

y∈R


We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

We multiply parentheses

-2y^2+y+(-2y-3)-4y=0

We get rid of parentheses

-2y^2+y-2y-4y-3=0

We add all the numbers together, and all the variables

-2y^2-5y-3=0

a = -2; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·(-2)·(-3)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*-2}=\frac{4}{-4}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*-2}=\frac{6}{-4}
=-1+1/2
$