(3/5x)+(7/2x)=1

Solution for (3/5x)+(7/2x)=1 equation:

(3/5x)+(7/2x)=1

We move all terms to the left:

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


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

3/5x+7/2x-1=0

We calculate fractions


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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

41x-1*10x^2=0

Wy multiply elements

-10x^2+41x=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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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