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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



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

(k+2)-(3k-2)

We get rid of parentheses

k-3k+2+2

We add all the numbers together, and all the variables

-2k+4

Back to the equation:

-(-2k+4)



We add all the numbers together, and all the variables

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

We get rid of parentheses

k+2k-4-4=0

We add all the numbers together, and all the variables

3k-8=0

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

3k=8

k=8/3

k=2+2/3

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