16/5y+5y=41/5y

Solution for 16/5y+5y=41/5y equation:

16/5y+5y=41/5y

We move all terms to the left:

16/5y+5y-(41/5y)=0


Domain of the equation: 5y!=0

y!=0/5

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

16/5y+5y-(+41/5y)=0

We add all the numbers together, and all the variables

5y+16/5y-(+41/5y)=0

We get rid of parentheses

5y+16/5y-41/5y=0

We multiply all the terms by the denominator


5y*5y+16-41=0

We add all the numbers together, and all the variables

5y*5y-25=0

Wy multiply elements

25y^2-25=0

a = 25; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·25·(-25)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-50}{2*25}=\frac{-50}{50}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+50}{2*25}=\frac{50}{50}
=1
$