1/3x+2/3x+4=x+2

Solution for 1/3x+2/3x+4=x+2 equation:

1/3x+2/3x+4=x+2

We move all terms to the left:

1/3x+2/3x+4-(x+2)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We get rid of parentheses

1/3x+2/3x-x-2+4=0

We multiply all the terms by the denominator


-x*3x-2*3x+4*3x+1+2=0

We add all the numbers together, and all the variables

-x*3x-2*3x+4*3x+3=0

Wy multiply elements

-3x^2-6x+12x+3=0

We add all the numbers together, and all the variables

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

a = -3; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·(-3)·3
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{2}}{2*-3}=\frac{-6-6\sqrt{2}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{2}}{2*-3}=\frac{-6+6\sqrt{2}}{-6}
$