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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