5(n+2)-5=4-(2n-5)

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Solution for 5(n+2)-5=4-(2n-5) equation: 



5(n+2)-5=4-(2n-5)

We move all terms to the left:

5(n+2)-5-(4-(2n-5))=0

We multiply parentheses

5n-(4-(2n-5))+10-5=0



We calculate terms in parentheses: -(4-(2n-5)), so:

4-(2n-5)

determiningTheFunctionDomain
-(2n-5)+4

We get rid of parentheses

-2n+5+4

We add all the numbers together, and all the variables

-2n+9

Back to the equation:

-(-2n+9)



We add all the numbers together, and all the variables

5n-(-2n+9)+5=0

We get rid of parentheses

5n+2n-9+5=0

We add all the numbers together, and all the variables

7n-4=0

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

7n=4

n=4/7

n=4/7

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