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

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



2(k+2)=3(6-k)
We move all terms to the left:
2(k+2)-(3(6-k))=0
We add all the numbers together, and all the variables
2(k+2)-(3(-1k+6))=0
We multiply parentheses
2k-(3(-1k+6))+4=0
We calculate terms in parentheses: -(3(-1k+6)), so:
3(-1k+6)
We multiply parentheses
-3k+18
Back to the equation:
-(-3k+18)
We get rid of parentheses
2k+3k-18+4=0
We add all the numbers together, and all the variables
5k-14=0
We move all terms containing k to the left, all other terms to the right
5k=14
k=14/5
k=2+4/5

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