Simplifying an expression

**mj.alawami** · February 21st 2009, 06:02 AM

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

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

**skeeter** · February 21st 2009, 06:04 AM

Originally Posted by mj.alawami

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

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

xyz is the LCD ...

$\frac{z(x-y)}{xyz} +\frac{y(x-z)}{xyz}-\frac{x(z-y)}{xyz}$

now finish

**ADARSH** · February 21st 2009, 06:07 AM

You can also follow

$ =\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

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

One more thing its always better to ask different questions in different threads

**Soroban** · February 21st 2009, 06:19 AM

Helli, mj.alawami!

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

The LCD is $xyz$

We must "convert" each fraction so they all have the LCD.
. . We multiply each fraction by an appropriate fraction.

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

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

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

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

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

**mj.alawami** · February 21st 2009, 06:40 AM

Originally Posted by skeeter

xyz is the LCD ...

$\frac{z(x-y)}{xyz} +\frac{y(x-z)}{xyz}-\frac{x(z-y)}{xyz}$

now finish

Attempt: [Is this correct

]
$ \frac{-2zy+2xy}{xyz}= \frac{y(-2z+2x)}{y(xz)}= \frac{-2z+2x}{xz} $

**skeeter** · February 21st 2009, 07:00 AM

Originally Posted by mj.alawami

Attempt: [Is this correct

]
$ \frac{-2zy+2xy}{xyz}= \frac{y(-2z+2x)}{y(xz)}= \frac{-2z+2x}{xz} $

look at Soroban's post ... is his final result the same as yours?

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