1/3y+10=1/9y

Solution for 1/3y+10=1/9y equation:

1/3y+10=1/9y

We move all terms to the left:

1/3y+10-(1/9y)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 9y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

1/3y-(+1/9y)+10=0

We get rid of parentheses

1/3y-1/9y+10=0

We calculate fractions


9y/27y^2+(-3y)/27y^2+10=0

We multiply all the terms by the denominator


9y+(-3y)+10*27y^2=0

Wy multiply elements

270y^2+9y+(-3y)=0

We get rid of parentheses

270y^2+9y-3y=0

We add all the numbers together, and all the variables

270y^2+6y=0

a = 270; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·270·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*270}=\frac{-12}{540}
=-1/45
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*270}=\frac{0}{540}
=0
$