3(c-3)+2c=-(c-5)+4

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Solution for 3(c-3)+2c=-(c-5)+4 equation:



3(c-3)+2c=-(c-5)+4
We move all terms to the left:
3(c-3)+2c-(-(c-5)+4)=0
We add all the numbers together, and all the variables
2c+3(c-3)-(-(c-5)+4)=0
We multiply parentheses
2c+3c-(-(c-5)+4)-9=0
We calculate terms in parentheses: -(-(c-5)+4), so:
-(c-5)+4
We get rid of parentheses
-c+5+4
We add all the numbers together, and all the variables
-1c+9
Back to the equation:
-(-1c+9)
We add all the numbers together, and all the variables
5c-(-1c+9)-9=0
We get rid of parentheses
5c+1c-9-9=0
We add all the numbers together, and all the variables
6c-18=0
We move all terms containing c to the left, all other terms to the right
6c=18
c=18/6
c=3

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