(6/7)k+(4/3)=1-(-2/3)k

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Solution for (6/7)k+(4/3)=1-(-2/3)k equation: 



(6/7)k+(4/3)=1-(-2/3)k

We move all terms to the left:

(6/7)k+(4/3)-(1-(-2/3)k)=0



Domain of the equation: 7)k!=0

k!=0/1

k!=0

k∈R





Domain of the equation: 3)k)!=0

k!=0/1

k!=0

k∈R



We add all the numbers together, and all the variables

(+6/7)k-(1-(-2/3)k)+(+4/3)=0

We multiply parentheses

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

We get rid of parentheses

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

We calculate fractions


6k^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:
$k=\frac{-b}{2a}=\frac{0}{12}=0$

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