2k+2/3=(k-3)/2-10

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Solution for 2k+2/3=(k-3)/2-10 equation: 



2k+2/3=(k-3)/2-10

We move all terms to the left:

2k+2/3-((k-3)/2-10)=0

We calculate fractions


2k+(-((k-3)*3)/()+()/()=0



We calculate terms in parentheses: +(-((k-3)*3)/()+()/(), so:

-((k-3)*3)/()+()/(

We add all the numbers together, and all the variables

-((k-3)*3)/()+1

We multiply all the terms by the denominator


-((k-3)*3)+1*()



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

(k-3)*3

We multiply parentheses

3k-9

Back to the equation:

-(3k-9)



We add all the numbers together, and all the variables

-(3k-9)

We get rid of parentheses

-3k+9

Back to the equation:

+(-3k+9)



We get rid of parentheses

2k-3k+9=0

We add all the numbers together, and all the variables

-1k+9=0

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

-k=-9

k=-9/-1

k=+9

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