9/(4x)+1/3=5/(2x)

Solution for 9/(4x)+1/3=5/(2x) equation:

9/(4x)+1/3=5/(2x)

We move all terms to the left:

9/(4x)+1/3-(5/(2x))=0


Domain of the equation: 4x!=0

x!=0/4

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

9/4x-(+5/2x)+1/3=0

We get rid of parentheses

9/4x-5/2x+1/3=0

We calculate fractions


16x^2/72x^2+162x/72x^2+(-180x)/72x^2=0

We multiply all the terms by the denominator


16x^2+162x+(-180x)=0

We get rid of parentheses

16x^2+162x-180x=0

We add all the numbers together, and all the variables

16x^2-18x=0

a = 16; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·16·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*16}=\frac{36}{32}
=1+1/8
$