48-(3c+4)=4(c+4)+c

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Solution for 48-(3c+4)=4(c+4)+c equation:



48-(3c+4)=4(c+4)+c
We move all terms to the left:
48-(3c+4)-(4(c+4)+c)=0
We get rid of parentheses
-3c-(4(c+4)+c)-4+48=0
We calculate terms in parentheses: -(4(c+4)+c), so:
4(c+4)+c
We add all the numbers together, and all the variables
c+4(c+4)
We multiply parentheses
c+4c+16
We add all the numbers together, and all the variables
5c+16
Back to the equation:
-(5c+16)
We add all the numbers together, and all the variables
-3c-(5c+16)+44=0
We get rid of parentheses
-3c-5c-16+44=0
We add all the numbers together, and all the variables
-8c+28=0
We move all terms containing c to the left, all other terms to the right
-8c=-28
c=-28/-8
c=3+1/2

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