6(c-1)-7=3(c+1)+5

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Solution for 6(c-1)-7=3(c+1)+5 equation: 



6(c-1)-7=3(c+1)+5

We move all terms to the left:

6(c-1)-7-(3(c+1)+5)=0

We multiply parentheses

6c-(3(c+1)+5)-6-7=0



We calculate terms in parentheses: -(3(c+1)+5), so:

3(c+1)+5

We multiply parentheses

3c+3+5

We add all the numbers together, and all the variables

3c+8

Back to the equation:

-(3c+8)



We add all the numbers together, and all the variables

6c-(3c+8)-13=0

We get rid of parentheses

6c-3c-8-13=0

We add all the numbers together, and all the variables

3c-21=0

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

3c=21

c=21/3

c=7

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