(13/2y)-(5/y)=10

Solution for (13/2y)-(5/y)=10 equation:

(13/2y)-(5/y)=10

We move all terms to the left:

(13/2y)-(5/y)-(10)=0


Domain of the equation: 2y)!=0

y!=0/1

y!=0

y∈R



Domain of the equation: y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

(+13/2y)-(+5/y)-10=0

We get rid of parentheses

13/2y-5/y-10=0

We calculate fractions


13y/2y^2+(-10y)/2y^2-10=0

We multiply all the terms by the denominator


13y+(-10y)-10*2y^2=0

Wy multiply elements

-20y^2+13y+(-10y)=0

We get rid of parentheses

-20y^2+13y-10y=0

We add all the numbers together, and all the variables

-20y^2+3y=0

a = -20; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-20)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-20}=\frac{-6}{-40}
=3/20
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-20}=\frac{0}{-40}
=0
$