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

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

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

We move all terms to the left:

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


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R



Domain of the equation: 3y-1)!=0

y∈R


We get rid of parentheses

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

We calculate fractions


(-27y)/135y^2+(-20y)/135y^2+(-20y)/135y^2+1=0

We multiply all the terms by the denominator


(-27y)+(-20y)+(-20y)+1*135y^2=0

Wy multiply elements

135y^2+(-27y)+(-20y)+(-20y)=0

We get rid of parentheses

135y^2-27y-20y-20y=0

We add all the numbers together, and all the variables

135y^2-67y=0

a = 135; b = -67; c = 0;
Δ = b2-4ac
Δ = -672-4·135·0
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-67)-67}{2*135}=\frac{0}{270}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-67)+67}{2*135}=\frac{134}{270}
=67/135
$