58-(2c+)=4(c+5)+c

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Solution for 58-(2c+)=4(c+5)+c equation:



58-(2c+)=4(c+5)+c
We move all terms to the left:
58-(2c+)-(4(c+5)+c)=0
We add all the numbers together, and all the variables
-(+2c)-(4(c+5)+c)+58=0
We get rid of parentheses
-2c-(4(c+5)+c)+58=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 get rid of parentheses
-2c-5c-20+58=0
We add all the numbers together, and all the variables
-7c+38=0
We move all terms containing c to the left, all other terms to the right
-7c=-38
c=-38/-7
c=5+3/7

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