(7/5y)+(5/y)=1

Solution for (7/5y)+(5/y)=1 equation:

(7/5y)+(5/y)=1

We move all terms to the left:

(7/5y)+(5/y)-(1)=0


Domain of the equation: 5y)!=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

(+7/5y)+(+5/y)-1=0

We get rid of parentheses

7/5y+5/y-1=0

We calculate fractions


7y/5y^2+25y/5y^2-1=0

We multiply all the terms by the denominator


7y+25y-1*5y^2=0

We add all the numbers together, and all the variables

32y-1*5y^2=0

Wy multiply elements

-5y^2+32y=0

a = -5; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·(-5)·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*-5}=\frac{-64}{-10}
=6+2/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*-5}=\frac{0}{-10}
=0
$