9/5x+83=4/9x+92

Solution for 9/5x+83=4/9x+92 equation:

9/5x+83=4/9x+92

We move all terms to the left:

9/5x+83-(4/9x+92)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 9x+92)!=0

x∈R


We get rid of parentheses

9/5x-4/9x-92+83=0

We calculate fractions


81x/45x^2+(-20x)/45x^2-92+83=0

We add all the numbers together, and all the variables

81x/45x^2+(-20x)/45x^2-9=0

We multiply all the terms by the denominator


81x+(-20x)-9*45x^2=0

Wy multiply elements

-405x^2+81x+(-20x)=0

We get rid of parentheses

-405x^2+81x-20x=0

We add all the numbers together, and all the variables

-405x^2+61x=0

a = -405; b = 61; c = 0;
Δ = b2-4ac
Δ = 612-4·(-405)·0
Δ = 3721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3721}=61$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(61)-61}{2*-405}=\frac{-122}{-810}
=61/405
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(61)+61}{2*-405}=\frac{0}{-810}
=0
$