1/3v+5=2/9v+7

Solution for 1/3v+5=2/9v+7 equation:

1/3v+5=2/9v+7

We move all terms to the left:

1/3v+5-(2/9v+7)=0


Domain of the equation: 3v!=0

v!=0/3

v!=0

v∈R



Domain of the equation: 9v+7)!=0

v∈R


We get rid of parentheses

1/3v-2/9v-7+5=0

We calculate fractions


9v/27v^2+(-6v)/27v^2-7+5=0

We add all the numbers together, and all the variables

9v/27v^2+(-6v)/27v^2-2=0

We multiply all the terms by the denominator


9v+(-6v)-2*27v^2=0

Wy multiply elements

-54v^2+9v+(-6v)=0

We get rid of parentheses

-54v^2+9v-6v=0

We add all the numbers together, and all the variables

-54v^2+3v=0

a = -54; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-54)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-54}=\frac{-6}{-108}
=1/18
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-54}=\frac{0}{-108}
=0
$