26-(3c+4)=3(c+5)+c

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Solution for 26-(3c+4)=3(c+5)+c equation: 



26-(3c+4)=3(c+5)+c

We move all terms to the left:

26-(3c+4)-(3(c+5)+c)=0

We get rid of parentheses

-3c-(3(c+5)+c)-4+26=0



We calculate terms in parentheses: -(3(c+5)+c), so:

3(c+5)+c

We add all the numbers together, and all the variables

c+3(c+5)

We multiply parentheses

c+3c+15

We add all the numbers together, and all the variables

4c+15

Back to the equation:

-(4c+15)



We add all the numbers together, and all the variables

-3c-(4c+15)+22=0

We get rid of parentheses

-3c-4c-15+22=0

We add all the numbers together, and all the variables

-7c+7=0

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

-7c=-7

c=-7/-7

c=1

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