(4/9x)+(1/5x)=58

Solution for (4/9x)+(1/5x)=58 equation:

(4/9x)+(1/5x)=58

We move all terms to the left:

(4/9x)+(1/5x)-(58)=0


Domain of the equation: 9x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+4/9x)+(+1/5x)-58=0

We get rid of parentheses

4/9x+1/5x-58=0

We calculate fractions


20x/45x^2+9x/45x^2-58=0

We multiply all the terms by the denominator


20x+9x-58*45x^2=0

We add all the numbers together, and all the variables

29x-58*45x^2=0

Wy multiply elements

-2610x^2+29x=0

a = -2610; b = 29; c = 0;
Δ = b2-4ac
Δ = 292-4·(-2610)·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-29}{2*-2610}=\frac{-58}{-5220}
=1/90
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+29}{2*-2610}=\frac{0}{-5220}
=0
$