40-(3c+)=4(c+5)+c

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Solution for 40-(3c+)=4(c+5)+c equation: 



40-(3c+)=4(c+5)+c

We move all terms to the left:

40-(3c+)-(4(c+5)+c)=0

We add all the numbers together, and all the variables

-(+3c)-(4(c+5)+c)+40=0

We get rid of parentheses

-3c-(4(c+5)+c)+40=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

-3c-5c-20+40=0

We add all the numbers together, and all the variables

-8c+20=0

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

-8c=-20

c=-20/-8

c=2+1/2

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