41-(2c+3)=4(1c+6)+c

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Solution for 41-(2c+3)=4(1c+6)+c equation: 



41-(2c+3)=4(1c+6)+c

We move all terms to the left:

41-(2c+3)-(4(1c+6)+c)=0

We add all the numbers together, and all the variables

-(2c+3)-(4(c+6)+c)+41=0

We get rid of parentheses

-2c-(4(c+6)+c)-3+41=0



We calculate terms in parentheses: -(4(c+6)+c), so:

4(c+6)+c

We add all the numbers together, and all the variables

c+4(c+6)

We multiply parentheses

c+4c+24

We add all the numbers together, and all the variables

5c+24

Back to the equation:

-(5c+24)



We add all the numbers together, and all the variables

-2c-(5c+24)+38=0

We get rid of parentheses

-2c-5c-24+38=0

We add all the numbers together, and all the variables

-7c+14=0

We move all terms containing c to the left, all other terms to the right

-7c=-14

c=-14/-7

c=+2

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