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?
Okay. Cross multiply the RHS giving:giving:
Now subtract RHS from both sides giving:
Cross multiply to give:
Factoring top and bottom 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 and
Not equal to 5 as this is an asymptote as is -5 from the pre-cancelled equation.
Last edited by gnitsuk (2006-09-22 03:03:07)
You lost a restriction, gnitsuk. x can not be -5.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Absoluely, corrected it. Thanks.
Last edited by gnitsuk (2006-09-22 02:49:48)
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