48-(3c+4)=2(c+5)+c

Solution for 48-(3c+4)=2(c+5)+c equation:

48-(3c+4)=2(c+5)+c

We move all terms to the left:

48-(3c+4)-(2(c+5)+c)=0

We get rid of parentheses

-3c-(2(c+5)+c)-4+48=0


We calculate terms in parentheses: -(2(c+5)+c), so:

2(c+5)+c

We add all the numbers together, and all the variables

c+2(c+5)

We multiply parentheses

c+2c+10

We add all the numbers together, and all the variables

3c+10

Back to the equation:

-(3c+10)


We add all the numbers together, and all the variables

-3c-(3c+10)+44=0

We get rid of parentheses

-3c-3c-10+44=0

We add all the numbers together, and all the variables

-6c+34=0

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

-6c=-34

c=-34/-6

c=5+2/3