1/12y+3=18+1/3y

Solution for 1/12y+3=18+1/3y equation:

1/12y+3=18+1/3y

We move all terms to the left:

1/12y+3-(18+1/3y)=0


Domain of the equation: 12y!=0

y!=0/12

y!=0

y∈R



Domain of the equation: 3y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

1/12y-(1/3y+18)+3=0

We get rid of parentheses

1/12y-1/3y-18+3=0

We calculate fractions


3y/36y^2+(-12y)/36y^2-18+3=0

We add all the numbers together, and all the variables

3y/36y^2+(-12y)/36y^2-15=0

We multiply all the terms by the denominator


3y+(-12y)-15*36y^2=0

Wy multiply elements

-540y^2+3y+(-12y)=0

We get rid of parentheses

-540y^2+3y-12y=0

We add all the numbers together, and all the variables

-540y^2-9y=0

a = -540; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-540)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-540}=\frac{0}{-1080}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-540}=\frac{18}{-1080}
=-1/60
$