(k-2+(4k/3)-2)=24

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Solution for (k-2+(4k/3)-2)=24 equation: 



(k-2+(4k/3)-2)=24

We move all terms to the left:

(k-2+(4k/3)-2)-(24)=0

We add all the numbers together, and all the variables

(k-2+(+4k/3)-2)-24=0

We multiply all the terms by the denominator


(k-2+(+4k-24*3)-2)=0



We calculate terms in parentheses: +(k-2+(+4k-24*3)-2), so:

k-2+(+4k-24*3)-2

determiningTheFunctionDomain
k+(+4k-24*3)-2-2

We add all the numbers together, and all the variables

k+(4k-72)-2-2

We add all the numbers together, and all the variables

k+(4k-72)-4

We get rid of parentheses

k+4k-72-4

We add all the numbers together, and all the variables

5k-76

Back to the equation:

+(5k-76)



We get rid of parentheses

5k-76=0

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

5k=76

k=76/5

k=15+1/5

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