3(k-2)-6=3k-(2k-1)

Solution for 3(k-2)-6=3k-(2k-1) equation:

3(k-2)-6=3k-(2k-1)

We move all terms to the left:

3(k-2)-6-(3k-(2k-1))=0

We multiply parentheses

3k-(3k-(2k-1))-6-6=0


We calculate terms in parentheses: -(3k-(2k-1)), so:

3k-(2k-1)

We get rid of parentheses

3k-2k+1

We add all the numbers together, and all the variables

k+1

Back to the equation:

-(k+1)


We add all the numbers together, and all the variables

3k-(k+1)-12=0

We get rid of parentheses

3k-k-1-12=0

We add all the numbers together, and all the variables

2k-13=0

We move all terms containing k to the left, all other terms to the right

2k=13

k=13/2

k=6+1/2