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3(x+1)=12+(x-1)
We move all terms to the left:
3(x+1)-(12+(x-1))=0
We multiply parentheses
3x-(12+(x-1))+3=0
We calculate terms in parentheses: -(12+(x-1)), so:We get rid of parentheses
12+(x-1)
determiningTheFunctionDomain (x-1)+12
We get rid of parentheses
x-1+12
We add all the numbers together, and all the variables
x+11
Back to the equation:
-(x+11)
3x-x-11+3=0
We add all the numbers together, and all the variables
2x-8=0
We move all terms containing x to the left, all other terms to the right
2x=8
x=8/2
x=4
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