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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2c+5c-(3c+6(-1c+2)+3)-20=0



We calculate terms in parentheses: -(3c+6(-1c+2)+3), so:

3c+6(-1c+2)+3

We multiply parentheses

3c-6c+12+3

We add all the numbers together, and all the variables

-3c+15

Back to the equation:

-(-3c+15)



We add all the numbers together, and all the variables

7c-(-3c+15)-20=0

We get rid of parentheses

7c+3c-15-20=0

We add all the numbers together, and all the variables

10c-35=0

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

10c=35

c=35/10

c=3+1/2

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