2k+4+6k=3(1k-2)

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Solution for 2k+4+6k=3(1k-2) equation: 



2k+4+6k=3(1k-2)

We move all terms to the left:

2k+4+6k-(3(1k-2))=0

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

8k-(3(k-2))+4=0



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

3(k-2)

We multiply parentheses

3k-6

Back to the equation:

-(3k-6)



We get rid of parentheses

8k-3k+6+4=0

We add all the numbers together, and all the variables

5k+10=0

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

5k=-10

k=-10/5

k=-2

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