56-(3c+4)=4(c-5)+c

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Solution for 56-(3c+4)=4(c-5)+c equation:



56-(3c+4)=4(c-5)+c
We move all terms to the left:
56-(3c+4)-(4(c-5)+c)=0
We get rid of parentheses
-3c-(4(c-5)+c)-4+56=0
We calculate terms in parentheses: -(4(c-5)+c), so:
4(c-5)+c
We add all the numbers together, and all the variables
c+4(c-5)
We multiply parentheses
c+4c-20
We add all the numbers together, and all the variables
5c-20
Back to the equation:
-(5c-20)
We add all the numbers together, and all the variables
-3c-(5c-20)+52=0
We get rid of parentheses
-3c-5c+20+52=0
We add all the numbers together, and all the variables
-8c+72=0
We move all terms containing c to the left, all other terms to the right
-8c=-72
c=-72/-8
c=+9

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