56-(3c+4)=4(c-5)+c

Solution for 56-(3c+4)=4(c-5)+c equation:

56-(3c+4)=4(c-5)+c

We move all terms to the left:

56-(3c+4)-(4(c-5)+c)=0

We get rid of parentheses

-3c-(4(c-5)+c)-4+56=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)+52=0

We get rid of parentheses

-3c-5c+20+52=0

We add all the numbers together, and all the variables

-8c+72=0

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

-8c=-72

c=-72/-8

c=+9