k-3(k-4)+4(1-3k)=2(4+k)-5(k-3)

Solution for k-3(k-4)+4(1-3k)=2(4+k)-5(k-3) equation:

k-3(k-4)+4(1-3k)=2(4+k)-5(k-3)

We move all terms to the left:

k-3(k-4)+4(1-3k)-(2(4+k)-5(k-3))=0

We add all the numbers together, and all the variables

k-3(k-4)+4(-3k+1)-(2(k+4)-5(k-3))=0

We multiply parentheses

k-3k-12k-(2(k+4)-5(k-3))+12+4=0


We calculate terms in parentheses: -(2(k+4)-5(k-3)), so:

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

We multiply parentheses

2k-5k+8+15

We add all the numbers together, and all the variables

-3k+23

Back to the equation:

-(-3k+23)


We add all the numbers together, and all the variables

-14k-(-3k+23)+16=0

We get rid of parentheses

-14k+3k-23+16=0

We add all the numbers together, and all the variables

-11k-7=0

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

-11k=7

k=7/-11

k=-7/11