5(k+2)-7=13-4(4-k)

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Solution for 5(k+2)-7=13-4(4-k) equation: 



5(k+2)-7=13-4(4-k)

We move all terms to the left:

5(k+2)-7-(13-4(4-k))=0

We add all the numbers together, and all the variables

5(k+2)-(13-4(-1k+4))-7=0

We multiply parentheses

5k-(13-4(-1k+4))+10-7=0



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

13-4(-1k+4)

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

We multiply parentheses

4k-16+13

We add all the numbers together, and all the variables

4k-3

Back to the equation:

-(4k-3)



We add all the numbers together, and all the variables

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

We get rid of parentheses

5k-4k+3+3=0

We add all the numbers together, and all the variables

k+6=0

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

k=-6

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