(7/4x)+(5/x)=2

Solution for (7/4x)+(5/x)=2 equation:

(7/4x)+(5/x)=2

We move all terms to the left:

(7/4x)+(5/x)-(2)=0


Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+7/4x)+(+5/x)-2=0

We get rid of parentheses

7/4x+5/x-2=0

We calculate fractions


7x/4x^2+20x/4x^2-2=0

We multiply all the terms by the denominator


7x+20x-2*4x^2=0

We add all the numbers together, and all the variables

27x-2*4x^2=0

Wy multiply elements

-8x^2+27x=0

a = -8; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·(-8)·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*-8}=\frac{-54}{-16}
=3+3/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*-8}=\frac{0}{-16}
=0
$