1/4y+4=1/5y

Solution for 1/4y+4=1/5y equation:

1/4y+4=1/5y

We move all terms to the left:

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


Domain of the equation: 4y!=0

y!=0/4

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

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

We get rid of parentheses

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

We calculate fractions


5y/20y^2+(-4y)/20y^2+4=0

We multiply all the terms by the denominator


5y+(-4y)+4*20y^2=0

Wy multiply elements

80y^2+5y+(-4y)=0

We get rid of parentheses

80y^2+5y-4y=0

We add all the numbers together, and all the variables

80y^2+y=0

a = 80; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·80·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*80}=\frac{-2}{160}
=-1/80
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*80}=\frac{0}{160}
=0
$