7/2y+3/5y=1

Solution for 7/2y+3/5y=1 equation:

7/2y+3/5y=1

We move all terms to the left:

7/2y+3/5y-(1)=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R


We calculate fractions


35y/10y^2+6y/10y^2-1=0

We multiply all the terms by the denominator


35y+6y-1*10y^2=0

We add all the numbers together, and all the variables

41y-1*10y^2=0

Wy multiply elements

-10y^2+41y=0

a = -10; b = 41; c = 0;
Δ = b2-4ac
Δ = 412-4·(-10)·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-41}{2*-10}=\frac{-82}{-20}
=4+1/10
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+41}{2*-10}=\frac{0}{-20}
=0
$