4(k-2)-8=3k-(2k-1)

Solution for 4(k-2)-8=3k-(2k-1) equation:

4(k-2)-8=3k-(2k-1)

We move all terms to the left:

4(k-2)-8-(3k-(2k-1))=0

We multiply parentheses

4k-(3k-(2k-1))-8-8=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

4k-(k+1)-16=0

We get rid of parentheses

4k-k-1-16=0

We add all the numbers together, and all the variables

3k-17=0

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

3k=17

k=17/3

k=5+2/3