1/9x+1/3x=56

Solution for 1/9x+1/3x=56 equation:

1/9x+1/3x=56

We move all terms to the left:

1/9x+1/3x-(56)=0


Domain of the equation: 9x!=0

x!=0/9

x!=0

x∈R



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We calculate fractions


3x/27x^2+9x/27x^2-56=0

We multiply all the terms by the denominator


3x+9x-56*27x^2=0

We add all the numbers together, and all the variables

12x-56*27x^2=0

Wy multiply elements

-1512x^2+12x=0

a = -1512; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-1512)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-1512}=\frac{-24}{-3024}
=1/126
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-1512}=\frac{0}{-3024}
=0
$