1-5(3k+2)=6-6(k+7)

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Solution for 1-5(3k+2)=6-6(k+7) equation: 



1-5(3k+2)=6-6(k+7)

We move all terms to the left:

1-5(3k+2)-(6-6(k+7))=0

We multiply parentheses

-15k-(6-6(k+7))-10+1=0



We calculate terms in parentheses: -(6-6(k+7)), so:

6-6(k+7)

determiningTheFunctionDomain
-6(k+7)+6

We multiply parentheses

-6k-42+6

We add all the numbers together, and all the variables

-6k-36

Back to the equation:

-(-6k-36)



We add all the numbers together, and all the variables

-15k-(-6k-36)-9=0

We get rid of parentheses

-15k+6k+36-9=0

We add all the numbers together, and all the variables

-9k+27=0

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

-9k=-27

k=-27/-9

k=+3

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