2/3v-5=16+3/4v

Solution for 2/3v-5=16+3/4v equation:

2/3v-5=16+3/4v

We move all terms to the left:

2/3v-5-(16+3/4v)=0


Domain of the equation: 3v!=0

v!=0/3

v!=0

v∈R



Domain of the equation: 4v)!=0

v!=0/1

v!=0

v∈R


We add all the numbers together, and all the variables

2/3v-(3/4v+16)-5=0

We get rid of parentheses

2/3v-3/4v-16-5=0

We calculate fractions


8v/12v^2+(-9v)/12v^2-16-5=0

We add all the numbers together, and all the variables

8v/12v^2+(-9v)/12v^2-21=0

We multiply all the terms by the denominator


8v+(-9v)-21*12v^2=0

Wy multiply elements

-252v^2+8v+(-9v)=0

We get rid of parentheses

-252v^2+8v-9v=0

We add all the numbers together, and all the variables

-252v^2-1v=0

a = -252; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-252)·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*-252}=\frac{0}{-504}
=0
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-252}=\frac{2}{-504}
=-1/252
$