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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