# Simplifying an expression

• Feb 21st 2009, 06:02 AM
mj.alawami
Simplifying 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 AM
skeeter
Quote:

Originally Posted by mj.alawami

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

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

xyz is the LCD ...

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

now finish
• Feb 21st 2009, 06:07 AM
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 AM
Soroban
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 AM
mj.alawami
Quote:

Originally Posted by skeeter
xyz is the LCD ...

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

now finish

Attempt: [Is this correct (Thinking) ]
$\displaystyle \frac{-2zy+2xy}{xyz}= \frac{y(-2z+2x)}{y(xz)}= \frac{-2z+2x}{xz}$
• Feb 21st 2009, 07:00 AM
skeeter
Quote:

Originally Posted by mj.alawami
Attempt: [Is this correct (Thinking) ]
$\displaystyle \frac{-2zy+2xy}{xyz}= \frac{y(-2z+2x)}{y(xz)}= \frac{-2z+2x}{xz}$

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