snow.fall15
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I've done this question but don't know what I did wrong!
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snow.fall15
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The questionName:  hlp.jpg
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Last edited by snow.fall15; 3 weeks ago
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snow.fall15
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What I did: Name:  WhatsApp Image 2020-09-26 at 12.48.20.jpeg
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snow.fall15
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The question: Name:  hlp.jpg
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Muttley79
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(Original post by snow.fall15)
The question: Name:  hlp.jpg
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When you multiplied both sides by (x + 3) you assumed it was positive ...
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RDKGames
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(Original post by snow.fall15)
The question: Name:  hlp.jpg
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To be more clear than the above user, whenever you have an inequality you should be aware that multiplying both sides by a negative number reverses the inequality.

E.g. 1 < 2, so mult both side by -1 and we should have  -1 > -2.

If you want to multiply through by (x+3) then you need to consider two cases; one where it is positive, and one where it is negative.

The usual approach, however, is to multiply through by (x+3)^2 instead [or generally, by the square of whatever the denominator is]. This is because (x+3)^2 is always positive, so you know that the inequality sign will not reverse.

Anyway, to help you with your question, you should notice (before doing anything) that the denominator on the LHS factorises nicely. We have;

\dfrac{x}{(x-3)(x+1)} \leq \dfrac{1}{x+3}

So, in order to clear the denominators, and preserve the inequality, you should multiply both sides through by (x-3)^2(x+1)^2(x+3)^2 to obtain

x(x-3)(x+2)(x+3)^2 \leq (x-3)^2(x+2)^2(x+3)

From there on in it should be straighforward to solve if you decide to move the RHS to LHS and fully factorise.
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DFranklin
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Aside from the multiplying by something -ve issues, there seem to be some major algebra errors in the top post (bad enough that combined with the layout and the bad image quality I'm not totally sure what the OP has actually done).
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