3(k-5)-5=2(k-4)-k

Solution for 3(k-5)-5=2(k-4)-k equation:

3(k-5)-5=2(k-4)-k

We move all terms to the left:

3(k-5)-5-(2(k-4)-k)=0

We multiply parentheses

3k-(2(k-4)-k)-15-5=0


We calculate terms in parentheses: -(2(k-4)-k), so:

2(k-4)-k

We add all the numbers together, and all the variables

-1k+2(k-4)

We multiply parentheses

-1k+2k-8

We add all the numbers together, and all the variables

k-8

Back to the equation:

-(k-8)


We add all the numbers together, and all the variables

3k-(k-8)-20=0

We get rid of parentheses

3k-k+8-20=0

We add all the numbers together, and all the variables

2k-12=0

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

2k=12

k=12/2

k=6