(1/x)+(4x/3x-1)=0

Solution for (1/x)+(4x/3x-1)=0 equation:

(1/x)+(4x/3x-1)=0


Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 3x-1)!=0

x∈R


We add all the numbers together, and all the variables

(+1/x)+(4x/3x-1)=0

We get rid of parentheses

1/x+4x/3x-1=0

We calculate fractions


4x^2/3x^2+3x/3x^2-1=0

We multiply all the terms by the denominator


4x^2+3x-1*3x^2=0

Wy multiply elements

4x^2-3x^2+3x=0

We add all the numbers together, and all the variables

x^2+3x=0

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

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*1}=\frac{0}{2}
=0
$