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10 Jun 2019, 03:32
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Hey guys, would highly appreciate it if someone could explain how can I solve the below question :please :please
[x^2-6x+9] / [18-6x]
[x^2-6x+9] / [18-6x]
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10 Jun 2019, 04:44
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kasravafayie
[x^2-6x+9] / [18-6x]
= [x^2-6x+9] /6[3-x]
= [x-3]^2 /[6(3-x)]
= [x-3]^2 /[(-6)(x-3)]
= [x-3]/[-6] (if x=/=3)
= [3-x]/[6]
=1/2-x/6
10 Jun 2019, 13:46
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kasravafayie
When you say 'same numbers with different signs', I assume you're talking about the way that this fraction reduces to:
\(\frac{(x-3)(x-3)}{6(3-x)}\)
That is, you have x-3 on the top, but 3-x on the bottom, and you're not sure how to factor. Right?
The answer is that you can rewrite 3-x as -(x-3), or -1 times (x-3).
\(\frac{(x-3)(x-3)}{-6(x-3)}\)
\(\frac{(x-3)}{-6}\)
When you've practiced this a bit, you won't need to explicitly think about that every single time! You'll just need to remember that (x-3)/(3-x) (or any fraction that's 'reversed' like that one) will reduce to -1.
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Thank you for understanding, and happy exploring!
