40-(3c-4)=2(c+6)+c

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Solution for 40-(3c-4)=2(c+6)+c equation:



40-(3c-4)=2(c+6)+c
We move all terms to the left:
40-(3c-4)-(2(c+6)+c)=0
We get rid of parentheses
-3c-(2(c+6)+c)+4+40=0
We calculate terms in parentheses: -(2(c+6)+c), so:
2(c+6)+c
We add all the numbers together, and all the variables
c+2(c+6)
We multiply parentheses
c+2c+12
We add all the numbers together, and all the variables
3c+12
Back to the equation:
-(3c+12)
We add all the numbers together, and all the variables
-3c-(3c+12)+44=0
We get rid of parentheses
-3c-3c-12+44=0
We add all the numbers together, and all the variables
-6c+32=0
We move all terms containing c to the left, all other terms to the right
-6c=-32
c=-32/-6
c=5+1/3

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