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

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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 $

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