(N-5)+3(n+2)=4(n-3)-1

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Solution for (N-5)+3(n+2)=4(n-3)-1 equation: 



(-5)+3(N+2)=4(N-3)-1

We move all terms to the left:

(-5)+3(N+2)-(4(N-3)-1)=0

We add all the numbers together, and all the variables

3(N+2)-(4(N-3)-1)-5=0

We multiply parentheses

3N-(4(N-3)-1)+6-5=0



We calculate terms in parentheses: -(4(N-3)-1), so:

4(N-3)-1

We multiply parentheses

4N-12-1

We add all the numbers together, and all the variables

4N-13

Back to the equation:

-(4N-13)



We add all the numbers together, and all the variables

3N-(4N-13)+1=0

We get rid of parentheses

3N-4N+13+1=0

We add all the numbers together, and all the variables

-1N+14=0

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

-N=-14

N=-14/-1

N=+14

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