4(c+5)+3c=2c+4(c-10)

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Solution for 4(c+5)+3c=2c+4(c-10) equation: 



4(c+5)+3c=2c+4(c-10)

We move all terms to the left:

4(c+5)+3c-(2c+4(c-10))=0

We add all the numbers together, and all the variables

3c+4(c+5)-(2c+4(c-10))=0

We multiply parentheses

3c+4c-(2c+4(c-10))+20=0



We calculate terms in parentheses: -(2c+4(c-10)), so:

2c+4(c-10)

We multiply parentheses

2c+4c-40

We add all the numbers together, and all the variables

6c-40

Back to the equation:

-(6c-40)



We add all the numbers together, and all the variables

7c-(6c-40)+20=0

We get rid of parentheses

7c-6c+40+20=0

We add all the numbers together, and all the variables

c+60=0

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

c=-60

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