13-(3c+2)=2(c+2)-5c

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Solution for 13-(3c+2)=2(c+2)-5c equation: 



13-(3c+2)=2(c+2)-5c

We move all terms to the left:

13-(3c+2)-(2(c+2)-5c)=0

We get rid of parentheses

-3c-(2(c+2)-5c)-2+13=0



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

2(c+2)-5c

We add all the numbers together, and all the variables

-5c+2(c+2)

We multiply parentheses

-5c+2c+4

We add all the numbers together, and all the variables

-3c+4

Back to the equation:

-(-3c+4)



We add all the numbers together, and all the variables

-3c-(-3c+4)+11=0

We get rid of parentheses

-3c+3c-4+11=0

We add all the numbers together, and all the variables

7!=0

There is no solution for this equation

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