3/3y+4=2/y-4

Solution for 3/3y+4=2/y-4 equation:

3/3y+4=2/y-4

We move all terms to the left:

3/3y+4-(2/y-4)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



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

y∈R


We get rid of parentheses

3/3y-2/y+4+4=0

We calculate fractions


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

We add all the numbers together, and all the variables

3y/3y^2+(-6y)/3y^2+8=0

We multiply all the terms by the denominator


3y+(-6y)+8*3y^2=0

Wy multiply elements

24y^2+3y+(-6y)=0

We get rid of parentheses

24y^2+3y-6y=0

We add all the numbers together, and all the variables

24y^2-3y=0

a = 24; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·24·0
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*24}=\frac{0}{48}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*24}=\frac{6}{48}
=1/8
$