__5-x___=-(x-5)=-1(underneath -1 is 1 x-5 = x-5 I hope this isn't confusing I was just trying to show how my problem looks on the page with the 5-x being over the x-5. It is a very simple problem but I am just asking about if my thought process is right by knowing that the two x-5's cancel each other and so we are just left with the -1 on the top and the regular 1 on the bottom

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__5-x___=-(x-5)=-1(underneath -1 is 1 x-5 = x-5 I hope this isn't confusing I was just trying to show how my problem looks on the page with the 5-x being over the x-5. It is a very simple problem but I am just asking about if my thought process is right by knowing that the two x-5's cancel each other and so we are just left with the -1 on the top and the regular 1 on the bottom

Mathematics
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you're right :), but you can simplify -1/1 to -1
yes, but you must remember, that x can not be 5, because 0/0 is not defined
that too!

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okay, thank you so much
I have another question about this kind of rational expression
I had an x^2-3x with a 2x-6 on the bottom of that. I understand that the x^2-3x breaks up into subsequent parentheses of (x+3x) (x-3x) do I just leave the 2x-6 alone as it is on the bottom?
That sounds strange, if I understood you correctly you must calculate \[ \frac{x^2 - 3x}{2x -6} = \frac{x\cdot(x-3)}{2\cdot(x-3)} = \frac{x}{2}\]

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