1/3x=2+x

Solution for 1/3x=2+x equation:

1/3x=2+x

We move all terms to the left:

1/3x-(2+x)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

1/3x-(x+2)=0

We get rid of parentheses

1/3x-x-2=0

We multiply all the terms by the denominator


-x*3x-2*3x+1=0

Wy multiply elements

-3x^2-6x+1=0

a = -3; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·(-3)·1
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{3}}{2*-3}=\frac{6-4\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{3}}{2*-3}=\frac{6+4\sqrt{3}}{-6}
$