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

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

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

We move all terms to the left:

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


Domain of the equation: x!=0

x∈R



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: x-2)!=0

x∈R


We get rid of parentheses

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

We calculate fractions


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

We multiply all the terms by the denominator


(-3x+1)+(-4x)+2*3x^2=0

Wy multiply elements

6x^2+(-3x+1)+(-4x)=0

We get rid of parentheses

6x^2-3x-4x+1=0

We add all the numbers together, and all the variables

6x^2-7x+1=0

a = 6; b = -7; c = +1;
Δ = b2-4ac
Δ = -72-4·6·1
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5}{2*6}=\frac{2}{12}
=1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5}{2*6}=\frac{12}{12}
=1
$