5k+3/6=13-(4-k)/9

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Solution for 5k+3/6=13-(4-k)/9 equation: 



5k+3/6=13-(4-k)/9

We move all terms to the left:

5k+3/6-(13-(4-k)/9)=0

We add all the numbers together, and all the variables

5k-(13-(-1k+4)/9)+3/6=0

We calculate fractions


5k+(-(13-(-1k+4)*6)/()+()/()=0



We calculate terms in parentheses: +(-(13-(-1k+4)*6)/()+()/(), so:

-(13-(-1k+4)*6)/()+()/(

We add all the numbers together, and all the variables

-(13-(-1k+4)*6)/()+1

We multiply all the terms by the denominator


-(13-(-1k+4)*6)+1*()



We calculate terms in parentheses: -(13-(-1k+4)*6), so:

13-(-1k+4)*6

determiningTheFunctionDomain
-(-1k+4)*6+13

We multiply parentheses

6k-24+13

We add all the numbers together, and all the variables

6k-11

Back to the equation:

-(6k-11)



We add all the numbers together, and all the variables

-(6k-11)

We get rid of parentheses

-6k+11

Back to the equation:

+(-6k+11)



We get rid of parentheses

5k-6k+11=0

We add all the numbers together, and all the variables

-1k+11=0

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

-k=-11

k=-11/-1

k=+11

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