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

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

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

We move all terms to the left:

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

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

We get rid of parentheses

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

We calculate fractions


9x/3x^2+4x/3x^2-1=0

We multiply all the terms by the denominator


9x+4x-1*3x^2=0

We add all the numbers together, and all the variables

13x-1*3x^2=0

Wy multiply elements

-3x^2+13x=0

a = -3; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-3)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-3}=\frac{-26}{-6}
=4+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-3}=\frac{0}{-6}
=0
$