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

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



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

We move all terms to the left:

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

We multiply parentheses

2k+12k-(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 add all the numbers together, and all the variables

14k-(3k-6)-4=0

We get rid of parentheses

14k-3k+6-4=0

We add all the numbers together, and all the variables

11k+2=0

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

11k=-2

k=-2/11

k=-2/11

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