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(x+3)+2=2+2+(x+1)
We move all terms to the left:
(x+3)+2-(2+2+(x+1))=0
We get rid of parentheses
x-(2+2+(x+1))+3+2=0
We calculate terms in parentheses: -(2+2+(x+1)), so:We add all the numbers together, and all the variables
2+2+(x+1)
determiningTheFunctionDomain (x+1)+2+2
We add all the numbers together, and all the variables
(x+1)+4
We get rid of parentheses
x+1+4
We add all the numbers together, and all the variables
x+5
Back to the equation:
-(x+5)
x-(x+5)+5=0
We get rid of parentheses
x-x-5+5=0
We add all the numbers together, and all the variables
=0
x=0/1
x=0
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