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

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



-40-(3c+4)=2(c+3)+c
We move all terms to the left:
-40-(3c+4)-(2(c+3)+c)=0
We get rid of parentheses
-3c-(2(c+3)+c)-4-40=0
We calculate terms in parentheses: -(2(c+3)+c), so:
2(c+3)+c
We add all the numbers together, and all the variables
c+2(c+3)
We multiply parentheses
c+2c+6
We add all the numbers together, and all the variables
3c+6
Back to the equation:
-(3c+6)
We add all the numbers together, and all the variables
-3c-(3c+6)-44=0
We get rid of parentheses
-3c-3c-6-44=0
We add all the numbers together, and all the variables
-6c-50=0
We move all terms containing c to the left, all other terms to the right
-6c=50
c=50/-6
c=-8+1/3

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