2(2k+3)=16-(2k-2)

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Solution for 2(2k+3)=16-(2k-2) equation: 



2(2k+3)=16-(2k-2)

We move all terms to the left:

2(2k+3)-(16-(2k-2))=0

We multiply parentheses

4k-(16-(2k-2))+6=0



We calculate terms in parentheses: -(16-(2k-2)), so:

16-(2k-2)

determiningTheFunctionDomain
-(2k-2)+16

We get rid of parentheses

-2k+2+16

We add all the numbers together, and all the variables

-2k+18

Back to the equation:

-(-2k+18)



We get rid of parentheses

4k+2k-18+6=0

We add all the numbers together, and all the variables

6k-12=0

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

6k=12

k=12/6

k=2

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