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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



We calculate terms in parentheses: -(4(c+5)), so:

4(c+5)

We multiply parentheses

4c+20

Back to the equation:

-(4c+20)



We add all the numbers together, and all the variables

-3c-(4c+20)+36=0

We get rid of parentheses

-3c-4c-20+36=0

We add all the numbers together, and all the variables

-7c+16=0

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

-7c=-16

c=-16/-7

c=2+2/7

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