5c-(2c+100)=4c-2(15+c)

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Solution for 5c-(2c+100)=4c-2(15+c) equation: 



5c-(2c+100)=4c-2(15+c)

We move all terms to the left:

5c-(2c+100)-(4c-2(15+c))=0

We add all the numbers together, and all the variables

5c-(2c+100)-(4c-2(c+15))=0

We get rid of parentheses

5c-2c-(4c-2(c+15))-100=0



We calculate terms in parentheses: -(4c-2(c+15)), so:

4c-2(c+15)

We multiply parentheses

4c-2c-30

We add all the numbers together, and all the variables

2c-30

Back to the equation:

-(2c-30)



We add all the numbers together, and all the variables

3c-(2c-30)-100=0

We get rid of parentheses

3c-2c+30-100=0

We add all the numbers together, and all the variables

c-70=0

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

c=70

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