31=2(n-2)-3(2n-5)

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Solution for 31=2(n-2)-3(2n-5) equation: 



31=2(n-2)-3(2n-5)

We move all terms to the left:

31-(2(n-2)-3(2n-5))=0



We calculate terms in parentheses: -(2(n-2)-3(2n-5)), so:

2(n-2)-3(2n-5)

We multiply parentheses

2n-6n-4+15

We add all the numbers together, and all the variables

-4n+11

Back to the equation:

-(-4n+11)



We get rid of parentheses

4n-11+31=0

We add all the numbers together, and all the variables

4n+20=0

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

4n=-20

n=-20/4

n=-5

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