((12k-27)/3)-3=3+2k

Solution for ((12k-27)/3)-3=3+2k equation:

((12k-27)/3)-3=3+2k

We move all terms to the left:

((12k-27)/3)-3-(3+2k)=0

We add all the numbers together, and all the variables

((12k-27)/3)-(2k+3)-3=0

We get rid of parentheses

((12k-27)/3)-2k-3-3=0

We multiply all the terms by the denominator


((12k-27)-2k*3)-3*3)-3*3)=0


We calculate terms in parentheses: +((12k-27)-2k*3), so:

(12k-27)-2k*3

Wy multiply elements

(12k-27)-6k

We get rid of parentheses

12k-6k-27

We add all the numbers together, and all the variables

6k-27

Back to the equation:

+(6k-27)


We add all the numbers together, and all the variables

(6k-27)=0

We get rid of parentheses

6k-27=0

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

6k=27

k=27/6

k=4+1/2