(7/3x)+(9/6x)=50/9

Solution for (7/3x)+(9/6x)=50/9 equation:

(7/3x)+(9/6x)=50/9

We move all terms to the left:

(7/3x)+(9/6x)-(50/9)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 6x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+7/3x)+(+9/6x)-(+50/9)=0

We get rid of parentheses

7/3x+9/6x-50/9=0

We calculate fractions


(-5400x^2)/1458x^2+3402x/1458x^2+2187x/1458x^2=0

We multiply all the terms by the denominator


(-5400x^2)+3402x+2187x=0

We add all the numbers together, and all the variables

(-5400x^2)+5589x=0

We get rid of parentheses

-5400x^2+5589x=0

a = -5400; b = 5589; c = 0;
Δ = b2-4ac
Δ = 55892-4·(-5400)·0
Δ = 31236921
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{31236921}=5589$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5589)-5589}{2*-5400}=\frac{-11178}{-10800}
=1+7/200
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5589)+5589}{2*-5400}=\frac{0}{-10800}
=0
$