(3k-2)=2-(k+2)

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



(3k-2)=2-(k+2)

We move all terms to the left:

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

We get rid of parentheses

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



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

2-(k+2)

determiningTheFunctionDomain
-(k+2)+2

We get rid of parentheses

-k-2+2

We add all the numbers together, and all the variables

-k

Back to the equation:

-(-k)



We add all the numbers together, and all the variables

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

We get rid of parentheses

3k+1k-2=0

We add all the numbers together, and all the variables

4k-2=0

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

4k=2

k=2/4

k=1/2

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