(2)/(3y)-5=(1)/(6y)-4

Solution for (2)/(3y)-5=(1)/(6y)-4 equation:

(2)/(3y)-5=(1)/(6y)-4

We move all terms to the left:

(2)/(3y)-5-((1)/(6y)-4)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 6y-4)!=0

y∈R


We get rid of parentheses

2/3y-1/6y+4-5=0

We calculate fractions


12y/18y^2+(-3y)/18y^2+4-5=0

We add all the numbers together, and all the variables

12y/18y^2+(-3y)/18y^2-1=0

We multiply all the terms by the denominator


12y+(-3y)-1*18y^2=0

Wy multiply elements

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

We get rid of parentheses

-18y^2+12y-3y=0

We add all the numbers together, and all the variables

-18y^2+9y=0

a = -18; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-18)·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*-18}=\frac{-18}{-36}
=1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-18}=\frac{0}{-36}
=0
$