5(k+2)=22+4(k-3)

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Solution for 5(k+2)=22+4(k-3) equation: 



5(k+2)=22+4(k-3)

We move all terms to the left:

5(k+2)-(22+4(k-3))=0

We multiply parentheses

5k-(22+4(k-3))+10=0



We calculate terms in parentheses: -(22+4(k-3)), so:

22+4(k-3)

determiningTheFunctionDomain
4(k-3)+22

We multiply parentheses

4k-12+22

We add all the numbers together, and all the variables

4k+10

Back to the equation:

-(4k+10)



We get rid of parentheses

5k-4k-10+10=0

We add all the numbers together, and all the variables

k=0

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