Am I wrong or are the answers off?

• Apr 26th 2013, 04:31 PM
thatsmessedup
Am I wrong or are the answers off?
They say the answers for 25, and 28 are both 0 but that's not what I get. Can someone please verify this. Thanks.
Attachment 28171
• Apr 26th 2013, 05:27 PM
Soroban
Re: Am I wrong or are the answers off?
Herllo, thatsmessedup!

You are right!

Quote:

They say the answers for 25 and 28 are both 0. . They are wrong!.

$\displaystyle 25.\;\frac{x}{x-1} - \frac{x}{1-x}$

$\displaystyle \frac{x}{x-1} - \frac{x}{1-x} \;=\;\frac{x}{x-1} - \frac{x}{\text{-}(x-1)} \;=\;\frac{x}{x-1} + \frac{x}{x-1} \;=\;\frac{2x}{x-1}$

Quote:

$\displaystyle 28.\;\frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)}$

$\displaystyle \frac{1}{(x-1)(x-2)}\cdot\frac{x-3}{x-3} + \frac{1}{(x-2)(x-3)}\cdot\frac{x-1}{x-1}$

. . . $\displaystyle =\;\frac{x-3}{(x-1)(x-2)(x-3)} + \frac{x-1}{(x-1)(x-2)(x-3)}$

. . . $\displaystyle =\;\frac{(x-3) + (x-1)}{(x-1)(x-2)(x-3)}$

. . . $\displaystyle =\;\frac{2x-4}{(x-1)(x-2)(x-3)}$

. . . $\displaystyle =\;\frac{2(x-2)}{(x-1)(x-2)(x-3)}$

. . . $\displaystyle =\;\frac{2}{(x-1)(x-3)}$
• Apr 26th 2013, 06:06 PM
Prove It
Re: Am I wrong or are the answers off?
Quote:

Originally Posted by Soroban
Herllo, thatsmessedup!

You are right!

$\displaystyle \frac{x}{x-1} - \frac{x}{1-x} \;=\;\frac{x}{x-1} - \frac{x}{\text{-}(x-1)} \;=\;\frac{x}{x-1} + \frac{x}{x-1} \;=\;\frac{2x}{x-1}$

$\displaystyle \frac{1}{(x-1)(x-2)}\cdot\frac{x-3}{x-3} + \frac{1}{(x-2)(x-3)}\cdot\frac{x-1}{x-1}$

. . . $\displaystyle =\;\frac{x-3}{(x-1)(x-2)(x-3)} + \frac{x-1}{(x-1)(x-2)(x-3)}$

. . . $\displaystyle =\;\frac{(x-3) + (x-1)}{(x-1)(x-2)(x-3)}$

. . . $\displaystyle =\;\frac{2x-4}{(x-1)(x-2)(x-3)}$

. . . $\displaystyle =\;\frac{2(x-2)}{(x-1)(x-2)(x-3)}$

. . . $\displaystyle =\;\frac{2}{(x-1)(x-3)}$

Of course, this is implying that \displaystyle \displaystyle \begin{align*} x \neq \left\{ 1, 2, 3 \right\} \end{align*}.
• Apr 26th 2013, 07:19 PM
Paze
Re: Am I wrong or are the answers off?
Quote:

Originally Posted by Prove It
Of course, this is implying that \displaystyle \displaystyle \begin{align*} x \neq \left\{ 1, 2, 3 \right\} \end{align*}.

Still means that the answer will never be 0, right?
• Apr 26th 2013, 07:30 PM
Gusbob
Re: Am I wrong or are the answers off?
Quote:

Originally Posted by Paze
Still means that the answer will never be 0, right?

Yes
• Apr 27th 2013, 09:38 AM
thatsmessedup
Re: Am I wrong or are the answers off?
I'm actually get this practice exam from here Glendale Community College : Sample Tests and Study Packets which is ridiculous!