2/3y+(9/y)=1

Solution for 2/3y+(9/y)=1 equation:

2/3y+(9/y)=1

We move all terms to the left:

2/3y+(9/y)-(1)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

2/3y+(+9/y)-1=0

We get rid of parentheses

2/3y+9/y-1=0

We calculate fractions


2y/3y^2+27y/3y^2-1=0

We multiply all the terms by the denominator


2y+27y-1*3y^2=0

We add all the numbers together, and all the variables

29y-1*3y^2=0

Wy multiply elements

-3y^2+29y=0

a = -3; b = 29; c = 0;
Δ = b2-4ac
Δ = 292-4·(-3)·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-29}{2*-3}=\frac{-58}{-6}
=9+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+29}{2*-3}=\frac{0}{-6}
=0
$