57=11+10+(2x)+(x+3)+(3x-3)

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Solution for 57=11+10+(2x)+(x+3)+(3x-3) equation: 



57=11+10+(2x)+(x+3)+(3x-3)

We move all terms to the left:

57-(11+10+(2x)+(x+3)+(3x-3))=0



We calculate terms in parentheses: -(11+10+2x+(x+3)+(3x-3)), so:

11+10+2x+(x+3)+(3x-3)

determiningTheFunctionDomain
2x+(x+3)+(3x-3)+11+10

We add all the numbers together, and all the variables

2x+(x+3)+(3x-3)+21

We get rid of parentheses

2x+x+3x+3-3+21

We add all the numbers together, and all the variables

6x+21

Back to the equation:

-(6x+21)



We get rid of parentheses

-6x-21+57=0

We add all the numbers together, and all the variables

-6x+36=0

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

-6x=-36

x=-36/-6

x=+6

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