9/5x+20=9/x+4=4/5

Solution for 9/5x+20=9/x+4=4/5 equation:

9/5x+20=9/x+4=4/5

We move all terms to the left:

9/5x+20-(9/x+4)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: x+4)!=0

x∈R


We get rid of parentheses

9/5x-9/x-4+20=0

We calculate fractions


9x/5x^2+(-45x)/5x^2-4+20=0

We add all the numbers together, and all the variables

9x/5x^2+(-45x)/5x^2+16=0

We multiply all the terms by the denominator


9x+(-45x)+16*5x^2=0

Wy multiply elements

80x^2+9x+(-45x)=0

We get rid of parentheses

80x^2+9x-45x=0

We add all the numbers together, and all the variables

80x^2-36x=0

a = 80; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·80·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*80}=\frac{0}{160}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*80}=\frac{72}{160}
=9/20
$