1(k+2)+(-5)=(3k-2)

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Solution for 1(k+2)+(-5)=(3k-2) equation: 



1(k+2)+(-5)=(3k-2)

We move all terms to the left:

1(k+2)+(-5)-((3k-2))=0

We add all the numbers together, and all the variables

1(k+2)-((3k-2))-5=0



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

(3k-2)

We get rid of parentheses

3k-2

Back to the equation:

-(3k-2)



We get rid of parentheses

1(k+2)-3k+2-5=0

We add all the numbers together, and all the variables

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

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

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

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