1/3y+7=1/9y-1

Solution for 1/3y+7=1/9y-1 equation:

1/3y+7=1/9y-1

We move all terms to the left:

1/3y+7-(1/9y-1)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 9y-1)!=0

y∈R


We get rid of parentheses

1/3y-1/9y+1+7=0

We calculate fractions


9y/27y^2+(-3y)/27y^2+1+7=0

We add all the numbers together, and all the variables

9y/27y^2+(-3y)/27y^2+8=0

We multiply all the terms by the denominator


9y+(-3y)+8*27y^2=0

Wy multiply elements

216y^2+9y+(-3y)=0

We get rid of parentheses

216y^2+9y-3y=0

We add all the numbers together, and all the variables

216y^2+6y=0

a = 216; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·216·0
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*216}=\frac{-12}{432}
=-1/36
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*216}=\frac{0}{432}
=0
$