-30+3c-7(c-3c)=4(c-5)+3c+27

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Solution for -30+3c-7(c-3c)=4(c-5)+3c+27 equation: 



-30+3c-7(c-3c)=4(c-5)+3c+27

We move all terms to the left:

-30+3c-7(c-3c)-(4(c-5)+3c+27)=0

We add all the numbers together, and all the variables

3c-7(-2c)-(4(c-5)+3c+27)-30=0

We multiply parentheses

3c+14c-(4(c-5)+3c+27)-30=0



We calculate terms in parentheses: -(4(c-5)+3c+27), so:

4(c-5)+3c+27

We add all the numbers together, and all the variables

3c+4(c-5)+27

We multiply parentheses

3c+4c-20+27

We add all the numbers together, and all the variables

7c+7

Back to the equation:

-(7c+7)



We add all the numbers together, and all the variables

17c-(7c+7)-30=0

We get rid of parentheses

17c-7c-7-30=0

We add all the numbers together, and all the variables

10c-37=0

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

10c=37

c=37/10

c=3+7/10

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