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

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



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

We move all terms to the left:

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

We multiply parentheses

3k-(2(k-2))-18=0



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

2(k-2)

We multiply parentheses

2k-4

Back to the equation:

-(2k-4)



We get rid of parentheses

3k-2k+4-18=0

We add all the numbers together, and all the variables

k-14=0

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

k=14

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