1/y+2=8/3y+1

Solution for 1/y+2=8/3y+1 equation:

1/y+2=8/3y+1

We move all terms to the left:

1/y+2-(8/3y+1)=0


Domain of the equation: y!=0

y∈R



Domain of the equation: 3y+1)!=0

y∈R


We get rid of parentheses

1/y-8/3y-1+2=0

We calculate fractions


3y/3y^2+(-8y)/3y^2-1+2=0

We add all the numbers together, and all the variables

3y/3y^2+(-8y)/3y^2+1=0

We multiply all the terms by the denominator


3y+(-8y)+1*3y^2=0

Wy multiply elements

3y^2+3y+(-8y)=0

We get rid of parentheses

3y^2+3y-8y=0

We add all the numbers together, and all the variables

3y^2-5y=0

a = 3; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·3·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*3}=\frac{0}{6}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*3}=\frac{10}{6}
=1+2/3
$