3x-3/x+3-3-3x/3-x=6-3x/x-3

Solution for 3x-3/x+3-3-3x/3-x=6-3x/x-3 equation:

3x-3/x+3-3-3x/3-x=6-3x/x-3

We move all terms to the left:

3x-3/x+3-3-3x/3-x-(6-3x/x-3)=0


Domain of the equation: x!=0

x∈R



Domain of the equation: x-3)!=0

x∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


(-3x^2)/3x+2x+(9x-3)/3x-3=0

We multiply all the terms by the denominator


(-3x^2)+2x*3x+(9x-3)-3*3x=0

Wy multiply elements

(-3x^2)+6x^2+(9x-3)-9x=0

We get rid of parentheses

-3x^2+6x^2+9x-9x-3=0

We add all the numbers together, and all the variables

3x^2-3=0

a = 3; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·3·(-3)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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