Can you please answer this question because i am getting lost ...

$\displaystyle \frac{x-y}{xy} +\frac{x-z}{xz}-\frac{z-y}{yz} $

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- Feb 21st 2009, 06:02 AMmj.alawamiSimplifying an expression
Can you please answer this question because i am getting lost ...

$\displaystyle \frac{x-y}{xy} +\frac{x-z}{xz}-\frac{z-y}{yz} $ - Feb 21st 2009, 06:04 AMskeeter
- Feb 21st 2009, 06:07 AMADARSH
You can also follow

http://www.mathhelpforum.com/math-he...01ff0387-1.gif

$\displaystyle

=\frac{x}{xy} - \frac{y}{xy} + \frac{x}{xz} +\frac{-z}{xz} - \frac{z}{yz} - \frac{-y}{yz}

$

Now see if anything gets cancelled in

$\displaystyle

=\frac{1}{y} - \frac{1}{x} + \frac{1}{z} +\frac{-1}{x} - \frac{1}{y} - \frac{-1}{z}

$

(Wait)One more thing its always better to ask different questions in different threads - Feb 21st 2009, 06:19 AMSoroban
Helli, mj.alawami!

Quote:

$\displaystyle \frac{x-y}{xy} +\frac{x-z}{xz}-\frac{z-y}{yz} $

The LCD is $\displaystyle xyz$

We must "convert" each fraction so they all have the LCD.

. . We multiply each fraction by an appropriate fraction.

$\displaystyle {\color{blue}\frac{z}{z}}\cdot\frac{x-y}{xy} + {\color{blue}\frac{y}{y}}\cdot\frac{x-z}{xz} - {\color{blue}\frac{x}{x}}\cdot\frac{z-y}{yz}$

. . $\displaystyle = \;\frac{z(x-y)}{xyz} + \frac{y(x-z)}{xyz} - \frac{x(z-y)}{xyz}$ . They have the same denominator.

. . $\displaystyle = \;\frac{z(x-y) + y(x-z) - x(z-y)}{xyz}$ . We can make one big fraction.

. . $\displaystyle = \;\frac{xz - yz + xy - yz - xz + xy}{xyz}$ . Simplify the numerator, factor, and reduce.

. . $\displaystyle = \;\frac{2xy - 2yz}{xyz} \;=\;\frac{2{\color{red}\rlap{/}}y(x-z)}{x{\color{red}\rlap{/}}yz} \;=\;\frac{2(x-z)}{xz} $

- Feb 21st 2009, 06:40 AMmj.alawami
- Feb 21st 2009, 07:00 AMskeeter