4(3+c)+c=c+44(3+c)+c=c+4

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Solution for 4(3+c)+c=c+44(3+c)+c=c+4 equation: 



4(3+c)+c=c+44(3+c)+c=c+4

We move all terms to the left:

4(3+c)+c-(c+44(3+c)+c)=0

We add all the numbers together, and all the variables

4(c+3)+c-(c+44(c+3)+c)=0

We add all the numbers together, and all the variables

c+4(c+3)-(c+44(c+3)+c)=0

We multiply parentheses

c+4c-(c+44(c+3)+c)+12=0



We calculate terms in parentheses: -(c+44(c+3)+c), so:

c+44(c+3)+c

We add all the numbers together, and all the variables

2c+44(c+3)

We multiply parentheses

2c+44c+132

We add all the numbers together, and all the variables

46c+132

Back to the equation:

-(46c+132)



We add all the numbers together, and all the variables

5c-(46c+132)+12=0

We get rid of parentheses

5c-46c-132+12=0

We add all the numbers together, and all the variables

-41c-120=0

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

-41c=120

c=120/-41

c=-2+38/41

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