3(c+5)=15+2(2c-1)

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Solution for 3(c+5)=15+2(2c-1) equation: 



3(c+5)=15+2(2c-1)

We move all terms to the left:

3(c+5)-(15+2(2c-1))=0

We multiply parentheses

3c-(15+2(2c-1))+15=0



We calculate terms in parentheses: -(15+2(2c-1)), so:

15+2(2c-1)

determiningTheFunctionDomain
2(2c-1)+15

We multiply parentheses

4c-2+15

We add all the numbers together, and all the variables

4c+13

Back to the equation:

-(4c+13)



We get rid of parentheses

3c-4c-13+15=0

We add all the numbers together, and all the variables

-1c+2=0

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

-c=-2

c=-2/-1

c=+2

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