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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2c+3c-(-(c-5)+4)-9=0



We calculate terms in parentheses: -(-(c-5)+4), so:

-(c-5)+4

We get rid of parentheses

-c+5+4

We add all the numbers together, and all the variables

-1c+9

Back to the equation:

-(-1c+9)



We add all the numbers together, and all the variables

5c-(-1c+9)-9=0

We get rid of parentheses

5c+1c-9-9=0

We add all the numbers together, and all the variables

6c-18=0

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

6c=18

c=18/6

c=3

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