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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3c+8c-(3(3c+4)+8)+20=0



We calculate terms in parentheses: -(3(3c+4)+8), so:

3(3c+4)+8

We multiply parentheses

9c+12+8

We add all the numbers together, and all the variables

9c+20

Back to the equation:

-(9c+20)



We add all the numbers together, and all the variables

11c-(9c+20)+20=0

We get rid of parentheses

11c-9c-20+20=0

We add all the numbers together, and all the variables

2c=0

c=0/2

c=0

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