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

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

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

We move all terms to the left:

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

We multiply parentheses

k+2k+3k-(4(k+3))+2+6=0


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

4(k+3)

We multiply parentheses

4k+12

Back to the equation:

-(4k+12)


We add all the numbers together, and all the variables

6k-(4k+12)+8=0

We get rid of parentheses

6k-4k-12+8=0

We add all the numbers together, and all the variables

2k-4=0

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

2k=4

k=4/2

k=2