(4/x)+(4/3x)=9

Solution for (4/x)+(4/3x)=9 equation:

(4/x)+(4/3x)=9

We move all terms to the left:

(4/x)+(4/3x)-(9)=0


Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+4/x)+(+4/3x)-9=0

We get rid of parentheses

4/x+4/3x-9=0

We calculate fractions


12x/3x^2+4x/3x^2-9=0

We multiply all the terms by the denominator


12x+4x-9*3x^2=0

We add all the numbers together, and all the variables

16x-9*3x^2=0

Wy multiply elements

-27x^2+16x=0

a = -27; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-27)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-27}=\frac{-32}{-54}
=16/27
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-27}=\frac{0}{-54}
=0
$