3y/2y-4=1/y-4

Solution for 3y/2y-4=1/y-4 equation:

3y/2y-4=1/y-4

We move all terms to the left:

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


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



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

y∈R


We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

We get rid of parentheses

3y^2-2y=0

a = 3; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·3·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*3}=\frac{0}{6}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*3}=\frac{4}{6}
=2/3
$