(5/3x)+(7/2x)=1

Solution for (5/3x)+(7/2x)=1 equation:

(5/3x)+(7/2x)=1

We move all terms to the left:

(5/3x)+(7/2x)-(1)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+5/3x)+(+7/2x)-1=0

We get rid of parentheses

5/3x+7/2x-1=0

We calculate fractions


10x/6x^2+21x/6x^2-1=0

We multiply all the terms by the denominator


10x+21x-1*6x^2=0

We add all the numbers together, and all the variables

31x-1*6x^2=0

Wy multiply elements

-6x^2+31x=0

a = -6; b = 31; c = 0;
Δ = b2-4ac
Δ = 312-4·(-6)·0
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-31}{2*-6}=\frac{-62}{-12}
=5+1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+31}{2*-6}=\frac{0}{-12}
=0
$