(1/6x)+(5/4x)=2

Solution for (1/6x)+(5/4x)=2 equation:

(1/6x)+(5/4x)=2

We move all terms to the left:

(1/6x)+(5/4x)-(2)=0


Domain of the equation: 6x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/6x)+(+5/4x)-2=0

We get rid of parentheses

1/6x+5/4x-2=0

We calculate fractions


4x/24x^2+30x/24x^2-2=0

We multiply all the terms by the denominator


4x+30x-2*24x^2=0

We add all the numbers together, and all the variables

34x-2*24x^2=0

Wy multiply elements

-48x^2+34x=0

a = -48; b = 34; c = 0;
Δ = b2-4ac
Δ = 342-4·(-48)·0
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-34}{2*-48}=\frac{-68}{-96}
=17/24
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+34}{2*-48}=\frac{0}{-96}
=0
$