6(k-3)-7=4k-(3k-2)

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Solution for 6(k-3)-7=4k-(3k-2) equation: 



6(k-3)-7=4k-(3k-2)

We move all terms to the left:

6(k-3)-7-(4k-(3k-2))=0

We multiply parentheses

6k-(4k-(3k-2))-18-7=0



We calculate terms in parentheses: -(4k-(3k-2)), so:

4k-(3k-2)

We get rid of parentheses

4k-3k+2

We add all the numbers together, and all the variables

k+2

Back to the equation:

-(k+2)



We add all the numbers together, and all the variables

6k-(k+2)-25=0

We get rid of parentheses

6k-k-2-25=0

We add all the numbers together, and all the variables

5k-27=0

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

5k=27

k=27/5

k=5+2/5

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