23-(2c+2)=2(c+2)+3c

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Solution for 23-(2c+2)=2(c+2)+3c equation: 



23-(2c+2)=2(c+2)+3c

We move all terms to the left:

23-(2c+2)-(2(c+2)+3c)=0

We get rid of parentheses

-2c-(2(c+2)+3c)-2+23=0



We calculate terms in parentheses: -(2(c+2)+3c), so:

2(c+2)+3c

We add all the numbers together, and all the variables

3c+2(c+2)

We multiply parentheses

3c+2c+4

We add all the numbers together, and all the variables

5c+4

Back to the equation:

-(5c+4)



We add all the numbers together, and all the variables

-2c-(5c+4)+21=0

We get rid of parentheses

-2c-5c-4+21=0

We add all the numbers together, and all the variables

-7c+17=0

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

-7c=-17

c=-17/-7

c=2+3/7

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