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**Mr. Big****Guest**

I have this inequation '2x / (x^2 - 25) <= 1 / (x + 5) + x / (3x - 15)' which I need to solve, and for my life I can't remember how it's done.

Could anyone please help me?

**gnitsuk****Member**- Registered: 2006-02-09
- Posts: 119

we need:

Okay. Cross multiply the RHS giving:

giving:Now subtract RHS from both sides giving:

Cross multiply to give:

Tidying gives:

Factoring top and bottom gives:

Calcelling gives:

Now a quotient of two functions will be less than zero when the two functions are of opposite sign. We can easily see that -(x-3) <= 0 when x >= 3 and also that 3(x-5) <= 0 when x <= 5 therefore the quotient of the two functions will be less than zero when 3 => x >= 5.

So the answer is:

or andNot equal to 5 as this is an asymptote as is -5 from the pre-cancelled equation.

Mitch.

*Last edited by gnitsuk (2006-09-22 03:03:07)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

You lost a restriction, gnitsuk. x can not be -5.

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**gnitsuk****Member**- Registered: 2006-02-09
- Posts: 119

Absoluely, corrected it. Thanks.

*Last edited by gnitsuk (2006-09-22 02:49:48)*

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**Mr. Big****Guest**

I'm such an idiot.

It turns out I DID remember how to do it and was doing it right, but got completely wrong answears due to a miscalculation.

So it's something like this:

2x / (x^2 - 25) <= 1 / (x + 5) + x / (3x - 15)

0 <= 1 / (x + 5) + x / 3(x - 5) - 2x / (x + 5)(x - 5)

0 <= (x^2 + 2x - 15) / 3(x + 5)(x - 5)

0 <= ((x - 3)(x + 5)) / 3(x + 5)(x - 5)

0 <= (x - 3) / 3(x - 5)

x <= 3 or x < 5, x != -5

Thanks anyway!

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