58-(2c+3)=4(c+5)=c

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Solution for 58-(2c+3)=4(c+5)=c equation:



58-(2c+3)=4(c+5)=c
We move all terms to the left:
58-(2c+3)-(4(c+5))=0
We get rid of parentheses
-2c-(4(c+5))-3+58=0
We calculate terms in parentheses: -(4(c+5)), so:
4(c+5)
We multiply parentheses
4c+20
Back to the equation:
-(4c+20)
We add all the numbers together, and all the variables
-2c-(4c+20)+55=0
We get rid of parentheses
-2c-4c-20+55=0
We add all the numbers together, and all the variables
-6c+35=0
We move all terms containing c to the left, all other terms to the right
-6c=-35
c=-35/-6
c=5+5/6

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