3(2k-5)=6(k-4)+9

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Solution for 3(2k-5)=6(k-4)+9 equation: 



3(2k-5)=6(k-4)+9

We move all terms to the left:

3(2k-5)-(6(k-4)+9)=0

We multiply parentheses

6k-(6(k-4)+9)-15=0



We calculate terms in parentheses: -(6(k-4)+9), so:

6(k-4)+9

We multiply parentheses

6k-24+9

We add all the numbers together, and all the variables

6k-15

Back to the equation:

-(6k-15)



We get rid of parentheses

6k-6k+15-15=0

We add all the numbers together, and all the variables

=0

k=0/1

k=0

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