Inequality

**OzzMan** · February 13th 2008, 10:28 PM

Solve the Inequality

$\frac{T-8}{3-T}<0$

I got $T<8\quad -T<-3\quad \Longrightarrow T>3$

But the books answers were completely opposte of what I got. I'm not exactly sure where I messed up. The book got $T>8\quad T<3$

Anyone know what I'm doing wrong?

**earboth** · February 13th 2008, 11:37 PM

Originally Posted by OzzMan

Solve the Inequality

$\frac{T-8}{3-T}<0$

I got $T<8\quad -T<-3\quad \Longrightarrow T>3$

But the books answers were completely opposte of what I got. I'm not exactly sure where I messed up. The book got $T>8\quad T<3$

Anyone know what I'm doing wrong?

Your inequality describes a negative fraction.
A fraction is negative if

- the numerator is negative (that means lower than zero) and the denominator is positive (that means greater than zero)

or

-- the numerator is positive (that means greater than zero) and the denominator is negative (that means lower than zero)

Transcribe these conditions into inequalities:

$T-8<0~\wedge~3-T>0~~\vee~~T-8>0~\wedge~3-T<0$

$\underbrace{T<8~\wedge~3>T}_{T < 3}~~\vee~~ \underbrace{T>8~\wedge~3<T}_{T>8}$

The final result is: $T < 3 ~\vee~ T > 8$

To check your calculations you can plug in some values for T to see if they give the right result.

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