1/3y-4=1/5y

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

1/3y-4=1/5y

We move all terms to the left:

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


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 5y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


5y/15y^2+(-3y)/15y^2-4=0

We multiply all the terms by the denominator


5y+(-3y)-4*15y^2=0

Wy multiply elements

-60y^2+5y+(-3y)=0

We get rid of parentheses

-60y^2+5y-3y=0

We add all the numbers together, and all the variables

-60y^2+2y=0

a = -60; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-60)·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*-60}=\frac{-4}{-120}
=1/30
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-60}=\frac{0}{-120}
=0
$