(1/6)n+(1/3)=(2/3)n,

Solution for (1/6)n+(1/3)=(2/3)n, equation:

(1/6)n+(1/3)=(2/3)n.

We move all terms to the left:

(1/6)n+(1/3)-((2/3)n.)=0


Domain of the equation: 6)n!=0

n!=0/1

n!=0

n∈R



Domain of the equation: 3)n.)!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

(+1/6)n-((+2/3)n.)+(+1/3)=0

We multiply parentheses

n^2-((+2/3)n.)+(+1/3)=0

We get rid of parentheses

n^2-((+2/3)n.)+1/3=0

We calculate fractions


n^2=0

a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

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