(2k-4)=3(k+2)-3k

Solution for (2k-4)=3(k+2)-3k equation:

(2k-4)=3(k+2)-3k

We move all terms to the left:

(2k-4)-(3(k+2)-3k)=0

We get rid of parentheses

2k-(3(k+2)-3k)-4=0


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

3(k+2)-3k

We add all the numbers together, and all the variables

-3k+3(k+2)

We multiply parentheses

-3k+3k+6

We add all the numbers together, and all the variables

6

Back to the equation:

-(6)


We add all the numbers together, and all the variables

2k-10=0

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

2k=10

k=10/2

k=5