4c+5(3-2c)=8c-3-(c+8)

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Solution for 4c+5(3-2c)=8c-3-(c+8) equation: 



4c+5(3-2c)=8c-3-(c+8)

We move all terms to the left:

4c+5(3-2c)-(8c-3-(c+8))=0

We add all the numbers together, and all the variables

4c+5(-2c+3)-(8c-3-(c+8))=0

We multiply parentheses

4c-10c-(8c-3-(c+8))+15=0



We calculate terms in parentheses: -(8c-3-(c+8)), so:

8c-3-(c+8)

determiningTheFunctionDomain
8c-(c+8)-3

We get rid of parentheses

8c-c-8-3

We add all the numbers together, and all the variables

7c-11

Back to the equation:

-(7c-11)



We add all the numbers together, and all the variables

-6c-(7c-11)+15=0

We get rid of parentheses

-6c-7c+11+15=0

We add all the numbers together, and all the variables

-13c+26=0

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

-13c=-26

c=-26/-13

c=+2

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