1/13-(2c+2)=2(c+2)+3c

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Solution for 1/13-(2c+2)=2(c+2)+3c equation: 



1/13-(2c+2)=2(c+2)+3c

We move all terms to the left:

1/13-(2c+2)-(2(c+2)+3c)=0

We get rid of parentheses

-2c-(2(c+2)+3c)-2+1/13=0

We multiply all the terms by the denominator


-2c*13-((2(c+2)+3c))*13+1-2*13=0

We add all the numbers together, and all the variables

-2c*13-((2(c+2)+3c))*13-25=0

Wy multiply elements

-26c-((2(c+2)+3c))*13-25=0

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

-26c-((2(c+2)+3c))*13=25

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