-(3n-2)-(2n-7)+3=-6(n-5)-(2n+2)+4

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Solution for -(3n-2)-(2n-7)+3=-6(n-5)-(2n+2)+4 equation:



-(3n-2)-(2n-7)+3=-6(n-5)-(2n+2)+4
We move all terms to the left:
-(3n-2)-(2n-7)+3-(-6(n-5)-(2n+2)+4)=0
We get rid of parentheses
-3n-2n-(-6(n-5)-(2n+2)+4)+2+7+3=0
We calculate terms in parentheses: -(-6(n-5)-(2n+2)+4), so:
-6(n-5)-(2n+2)+4
We multiply parentheses
-6n-(2n+2)+30+4
We get rid of parentheses
-6n-2n-2+30+4
We add all the numbers together, and all the variables
-8n+32
Back to the equation:
-(-8n+32)
We add all the numbers together, and all the variables
-5n-(-8n+32)+12=0
We get rid of parentheses
-5n+8n-32+12=0
We add all the numbers together, and all the variables
3n-20=0
We move all terms containing n to the left, all other terms to the right
3n=20
n=20/3
n=6+2/3

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