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(n+5)-(-2n+3)=(2n-5)+(n+4)
We move all terms to the left:
(n+5)-(-2n+3)-((2n-5)+(n+4))=0
We get rid of parentheses
n+2n-((2n-5)+(n+4))+5-3=0
We calculate terms in parentheses: -((2n-5)+(n+4)), so:We add all the numbers together, and all the variables
(2n-5)+(n+4)
We get rid of parentheses
2n+n-5+4
We add all the numbers together, and all the variables
3n-1
Back to the equation:
-(3n-1)
3n-(3n-1)+2=0
We get rid of parentheses
3n-3n+1+2=0
We add all the numbers together, and all the variables
3!=0
There is no solution for this equation
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