1/3x+1/3x+x+x=48

Solution for 1/3x+1/3x+x+x=48 equation:

1/3x+1/3x+x+x=48

We move all terms to the left:

1/3x+1/3x+x+x-(48)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

2x+1/3x+1/3x-48=0

We multiply all the terms by the denominator


2x*3x-48*3x+1+1=0

We add all the numbers together, and all the variables

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

Wy multiply elements

6x^2-144x+2=0

a = 6; b = -144; c = +2;
Δ = b2-4ac
Δ = -1442-4·6·2
Δ = 20688
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:

$\sqrt{\Delta}=\sqrt{20688}=\sqrt{16*1293}=\sqrt{16}*\sqrt{1293}=4\sqrt{1293}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-144)-4\sqrt{1293}}{2*6}=\frac{144-4\sqrt{1293}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-144)+4\sqrt{1293}}{2*6}=\frac{144+4\sqrt{1293}}{12}
$