2c-(4-4c)=3(7-2c)-15

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Solution for 2c-(4-4c)=3(7-2c)-15 equation: 



2c-(4-4c)=3(7-2c)-15

We move all terms to the left:

2c-(4-4c)-(3(7-2c)-15)=0

We add all the numbers together, and all the variables

2c-(-4c+4)-(3(-2c+7)-15)=0

We get rid of parentheses

2c+4c-(3(-2c+7)-15)-4=0



We calculate terms in parentheses: -(3(-2c+7)-15), so:

3(-2c+7)-15

We multiply parentheses

-6c+21-15

We add all the numbers together, and all the variables

-6c+6

Back to the equation:

-(-6c+6)



We add all the numbers together, and all the variables

6c-(-6c+6)-4=0

We get rid of parentheses

6c+6c-6-4=0

We add all the numbers together, and all the variables

12c-10=0

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

12c=10

c=10/12

c=5/6

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