2/3x+x=16-x

Solution for 2/3x+x=16-x equation:

2/3x+x=16-x

We move all terms to the left:

2/3x+x-(16-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

2/3x+x-(-1x+16)=0

We add all the numbers together, and all the variables

x+2/3x-(-1x+16)=0

We get rid of parentheses

x+2/3x+1x-16=0

We multiply all the terms by the denominator


x*3x+1x*3x-16*3x+2=0

Wy multiply elements

3x^2+3x^2-48x+2=0

We add all the numbers together, and all the variables

6x^2-48x+2=0

a = 6; b = -48; c = +2;
Δ = b2-4ac
Δ = -482-4·6·2
Δ = 2256
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:

$\sqrt{\Delta}=\sqrt{2256}=\sqrt{16*141}=\sqrt{16}*\sqrt{141}=4\sqrt{141}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-4\sqrt{141}}{2*6}=\frac{48-4\sqrt{141}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+4\sqrt{141}}{2*6}=\frac{48+4\sqrt{141}}{12}
$