1/3y-4=1/6y+5

Solution for 1/3y-4=1/6y+5 equation:

1/3y-4=1/6y+5

We move all terms to the left:

1/3y-4-(1/6y+5)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 6y+5)!=0

y∈R


We get rid of parentheses

1/3y-1/6y-5-4=0

We calculate fractions


6y/18y^2+(-3y)/18y^2-5-4=0

We add all the numbers together, and all the variables

6y/18y^2+(-3y)/18y^2-9=0

We multiply all the terms by the denominator


6y+(-3y)-9*18y^2=0

Wy multiply elements

-162y^2+6y+(-3y)=0

We get rid of parentheses

-162y^2+6y-3y=0

We add all the numbers together, and all the variables

-162y^2+3y=0

a = -162; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-162)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-162}=\frac{-6}{-324}
=1/54
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-162}=\frac{0}{-324}
=0
$