(3/7v)-(2/5v)=1

Solution for (3/7v)-(2/5v)=1 equation:

(3/7v)-(2/5v)=1

We move all terms to the left:

(3/7v)-(2/5v)-(1)=0


Domain of the equation: 7v)!=0

v!=0/1

v!=0

v∈R



Domain of the equation: 5v)!=0

v!=0/1

v!=0

v∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

3/7v-2/5v-1=0

We calculate fractions


15v/35v^2+(-14v)/35v^2-1=0

We multiply all the terms by the denominator


15v+(-14v)-1*35v^2=0

Wy multiply elements

-35v^2+15v+(-14v)=0

We get rid of parentheses

-35v^2+15v-14v=0

We add all the numbers together, and all the variables

-35v^2+v=0

a = -35; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-35)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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