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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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



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

3(k-4)

We multiply parentheses

3k-12

Back to the equation:

-(3k-12)



We add all the numbers together, and all the variables

7k-(3k-12)-8=0

We get rid of parentheses

7k-3k+12-8=0

We add all the numbers together, and all the variables

4k+4=0

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

4k=-4

k=-4/4

k=-1

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