5(c-4)+4=3+c-19

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Solution for 5(c-4)+4=3+c-19 equation: 



5(c-4)+4=3+c-19

We move all terms to the left:

5(c-4)+4-(3+c-19)=0

We add all the numbers together, and all the variables

5(c-4)-(c-16)+4=0

We multiply parentheses

5c-(c-16)-20+4=0

We get rid of parentheses

5c-c+16-20+4=0

We add all the numbers together, and all the variables

4c=0

c=0/4

c=0

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