2/3(k-3)=2/3(2k+1)

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

2/3(k-3)=2/3(2k+1)

We move all terms to the left:

2/3(k-3)-(2/3(2k+1))=0


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

k∈R



Domain of the equation: 3(2k+1))!=0

k∈R


We calculate fractions


(6k2/(3(k-3)*3(2k+1)))+(-6kk/(3(k-3)*3(2k+1)))=0


We calculate terms in parentheses: +(6k2/(3(k-3)*3(2k+1))), so:

6k2/(3(k-3)*3(2k+1))

We multiply all the terms by the denominator


6k2

We add all the numbers together, and all the variables

6k^2

Back to the equation:

+(6k^2)



We calculate terms in parentheses: +(-6kk/(3(k-3)*3(2k+1))), so:

-6kk/(3(k-3)*3(2k+1))

We multiply all the terms by the denominator


-6kk

Back to the equation:

+(-6kk)


We get rid of parentheses

6k^2-6kk=0