2/3(2-3x)=1/9(2x+1)

Solution for 2/3(2-3x)=1/9(2x+1) equation:

2/3(2-3x)=1/9(2x+1)

We move all terms to the left:

2/3(2-3x)-(1/9(2x+1))=0


Domain of the equation: 3(2-3x)!=0

x∈R



Domain of the equation: 9(2x+1))!=0

x∈R


We add all the numbers together, and all the variables

2/3(-3x+2)-(1/9(2x+1))=0

We calculate fractions


(18x2/(3(-3x+2)*9(2x+1)))+(-3x0/(3(-3x+2)*9(2x+1)))=0


We calculate terms in parentheses: +(18x2/(3(-3x+2)*9(2x+1))), so:

18x2/(3(-3x+2)*9(2x+1))

We multiply all the terms by the denominator


18x2

We add all the numbers together, and all the variables

18x^2

Back to the equation:

+(18x^2)



We calculate terms in parentheses: +(-3x0/(3(-3x+2)*9(2x+1))), so:

-3x0/(3(-3x+2)*9(2x+1))

We multiply all the terms by the denominator


-3x0

We add all the numbers together, and all the variables

-3x

Back to the equation:

+(-3x)


We get rid of parentheses

18x^2-3x=0

a = 18; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·18·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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