48-(3c+40)=4(c+5)+c

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Solution for 48-(3c+40)=4(c+5)+c equation: 



48-(3c+40)=4(c+5)+c

We move all terms to the left:

48-(3c+40)-(4(c+5)+c)=0

We get rid of parentheses

-3c-(4(c+5)+c)-40+48=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 add all the numbers together, and all the variables

-3c-(5c+20)+8=0

We get rid of parentheses

-3c-5c-20+8=0

We add all the numbers together, and all the variables

-8c-12=0

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

-8c=12

c=12/-8

c=-1+1/2

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