1/6x+1/3x=18

Solution for 1/6x+1/3x=18 equation:

1/6x+1/3x=18

We move all terms to the left:

1/6x+1/3x-(18)=0


Domain of the equation: 6x!=0

x!=0/6

x!=0

x∈R



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We calculate fractions


3x/18x^2+6x/18x^2-18=0

We multiply all the terms by the denominator


3x+6x-18*18x^2=0

We add all the numbers together, and all the variables

9x-18*18x^2=0

Wy multiply elements

-324x^2+9x=0

a = -324; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-324)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-324}=\frac{-18}{-648}
=1/36
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-324}=\frac{0}{-648}
=0
$