1/y=16/3y+9

Solution for 1/y=16/3y+9 equation:

1/y=16/3y+9

We move all terms to the left:

1/y-(16/3y+9)=0


Domain of the equation: y!=0

y∈R



Domain of the equation: 3y+9)!=0

y∈R


We get rid of parentheses

1/y-16/3y-9=0

We calculate fractions


3y/3y^2+(-16y)/3y^2-9=0

We multiply all the terms by the denominator


3y+(-16y)-9*3y^2=0

Wy multiply elements

-27y^2+3y+(-16y)=0

We get rid of parentheses

-27y^2+3y-16y=0

We add all the numbers together, and all the variables

-27y^2-13y=0

a = -27; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·(-27)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*-27}=\frac{0}{-54}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*-27}=\frac{26}{-54}
=-13/27
$