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

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

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

We move all terms to the left:

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


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 7x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


21x/35x^2+10x/35x^2-1=0

We multiply all the terms by the denominator


21x+10x-1*35x^2=0

We add all the numbers together, and all the variables

31x-1*35x^2=0

Wy multiply elements

-35x^2+31x=0

a = -35; b = 31; c = 0;
Δ = b2-4ac
Δ = 312-4·(-35)·0
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-31}{2*-35}=\frac{-62}{-70}
=31/35
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+31}{2*-35}=\frac{0}{-70}
=0
$