9(2k+3)+1=11(k-5)

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Solution for 9(2k+3)+1=11(k-5) equation: 



9(2k+3)+1=11(k-5)

We move all terms to the left:

9(2k+3)+1-(11(k-5))=0

We multiply parentheses

18k-(11(k-5))+27+1=0



We calculate terms in parentheses: -(11(k-5)), so:

11(k-5)

We multiply parentheses

11k-55

Back to the equation:

-(11k-55)



We add all the numbers together, and all the variables

18k-(11k-55)+28=0

We get rid of parentheses

18k-11k+55+28=0

We add all the numbers together, and all the variables

7k+83=0

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

7k=-83

k=-83/7

k=-11+6/7

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