7/3x+1/3x=3x+5/3x

Solution for 7/3x+1/3x=3x+5/3x equation:

7/3x+1/3x=3x+5/3x

We move all terms to the left:

7/3x+1/3x-(3x+5/3x)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

7/3x+1/3x-(+3x+5/3x)=0

We get rid of parentheses

7/3x+1/3x-3x-5/3x=0

We multiply all the terms by the denominator


-3x*3x+7+1-5=0

We add all the numbers together, and all the variables

-3x*3x+3=0

Wy multiply elements

-9x^2+3=0

a = -9; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-9)·3
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*-9}=\frac{0-6\sqrt{3}}{-18}
=-\frac{6\sqrt{3}}{-18}
=-\frac{\sqrt{3}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*-9}=\frac{0+6\sqrt{3}}{-18}
=\frac{6\sqrt{3}}{-18}
=\frac{\sqrt{3}}{-3}
$