3c-7(-c+4)=2(c-5)+6c

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Solution for 3c-7(-c+4)=2(c-5)+6c equation: 



3c-7(-c+4)=2(c-5)+6c

We move all terms to the left:

3c-7(-c+4)-(2(c-5)+6c)=0

We add all the numbers together, and all the variables

3c-7(-1c+4)-(2(c-5)+6c)=0

We multiply parentheses

3c+7c-(2(c-5)+6c)-28=0



We calculate terms in parentheses: -(2(c-5)+6c), so:

2(c-5)+6c

We add all the numbers together, and all the variables

6c+2(c-5)

We multiply parentheses

6c+2c-10

We add all the numbers together, and all the variables

8c-10

Back to the equation:

-(8c-10)



We add all the numbers together, and all the variables

10c-(8c-10)-28=0

We get rid of parentheses

10c-8c+10-28=0

We add all the numbers together, and all the variables

2c-18=0

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

2c=18

c=18/2

c=9

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