1/9(2x+16)=1/3(2x-4)

Solution for 1/9(2x+16)=1/3(2x-4) equation:

1/9(2x+16)=1/3(2x-4)

We move all terms to the left:

1/9(2x+16)-(1/3(2x-4))=0


Domain of the equation: 9(2x+16)!=0

x∈R



Domain of the equation: 3(2x-4))!=0

x∈R


We calculate fractions


(3x2/(9(2x+16)*3(2x-4)))+(-9x2/(9(2x+16)*3(2x-4)))=0


We calculate terms in parentheses: +(3x2/(9(2x+16)*3(2x-4))), so:

3x2/(9(2x+16)*3(2x-4))

We multiply all the terms by the denominator


3x2

We add all the numbers together, and all the variables

3x^2

Back to the equation:

+(3x^2)



We calculate terms in parentheses: +(-9x2/(9(2x+16)*3(2x-4))), so:

-9x2/(9(2x+16)*3(2x-4))

We multiply all the terms by the denominator


-9x2

We add all the numbers together, and all the variables

-9x^2

Back to the equation:

+(-9x^2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

3x^2-9x^2=0

We add all the numbers together, and all the variables

-6x^2=0

a = -6; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-6)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{-12}=0$