Factoring and canceling variables a inequalitiy question full points?

Question

say you have something like 3-4x+x^2/(x^2) can you just cancel out the x^@ from top botom now so left with 3-4x?

how would you solve 1/(3-x)>=(1-x)/(x^2)

Answer 1

First question: no.

Your expression is equivalent to 3/x^2 - 4x/x^2 + x^2/x^2.

(Do this sum and you'll see how where your expression came from.)

So you can only do a little canceling, say to get 3/x^2 - 4/x + 1, but the original is probably as simplified as you could get it.

Second:

1/(3 - x) >= (1 - x)/x^2

cross multiply

x^2 >= (3 - x)(1 - x)

x^2 >= x^2 - 4x + 3

0 >= 4x + 3

x <= -3/4  (but I would test a couple values around -3/4 and make sure the original inequality holds)

Answer 2

No...

1 >= (3-4x-x^2)/x^2

1 >= (3/x^2) -(4x/x^2) - (x^2/x^2)

1 >= (3/x^2) - (4x/x^2) - 1

+1                             +1

2 >= (3 - 4x) / x^2

That's as far as I've gotten. I'm sure I'm missing something. But, I do know you can't just cancel it out. :-/

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I figured it out (then hit refresh only to find the guy below me beat me to it, so I'm not gonna keep going. haha)