60-(2c+3)=4(c+7)+c

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Solution for 60-(2c+3)=4(c+7)+c equation:



60-(2c+3)=4(c+7)+c
We move all terms to the left:
60-(2c+3)-(4(c+7)+c)=0
We get rid of parentheses
-2c-(4(c+7)+c)-3+60=0
We calculate terms in parentheses: -(4(c+7)+c), so:
4(c+7)+c
We add all the numbers together, and all the variables
c+4(c+7)
We multiply parentheses
c+4c+28
We add all the numbers together, and all the variables
5c+28
Back to the equation:
-(5c+28)
We add all the numbers together, and all the variables
-2c-(5c+28)+57=0
We get rid of parentheses
-2c-5c-28+57=0
We add all the numbers together, and all the variables
-7c+29=0
We move all terms containing c to the left, all other terms to the right
-7c=-29
c=-29/-7
c=4+1/7

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