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15=5/3(x+12
We move all terms to the left:
15-(5/3(x+12)=0
Domain of the equation: 3(x+12)+15!=0We multiply all the terms by the denominator
We move all terms containing x to the left, all other terms to the right
3(x+12)!=-15
x∈R
-(5=0
We add all the numbers together, and all the variables
=0
x=0/1
x=0
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