• August 15th 2008, 08:51 AM
cmf0106
I need some help understanding whats going on here.

Description:
When adding fractions with variables in one or more denominators, the LCD will have each variable to its highest power as a factor.
$\frac{13}{xy^2}-\frac{6}{yz} = \frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z} =\frac{13z-6xy}{xy^2z}$

Can someone please help me through this process, I understand picking out the highest exponent of each variable but after that its unclear. Also how do you know to arrange the expression? For instance $\frac{13}{xy^2}*\frac{z}{z}$ How would I know that the $\frac{z}{z}$ gets matched up with the $\frac{13}{xy^2}$ and not on the other end of the expression with the $\frac{6xy}{xy^2z}$?

Many Thanks
• August 15th 2008, 09:10 AM
masters
Quote:

Originally Posted by cmf0106
I need some help understanding whats going on here.

Description:
When adding fractions with variables in one or more denominators, the LCD will have each variable to its highest power as a factor.
$\frac{13}{xy^2}-\frac{6}{yz} = \frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z} =\frac{13z-6xy}{xy^2z}$

Can someone please help me through this process, I understand picking out the highest exponent of each variable but after that its unclear. Also how do you know to arrange the expression? For instance $\frac{13}{xy^2}*\frac{z}{z}$ How would I know that the $\frac{z}{z}$ gets matched up with the $\frac{13}{xy^2}$ and not on the other end of the expression with the $\frac{6xy}{xy^2z}$?

Many Thanks

Let me introduce a whole different approach. If you don't like it, you can still pursue the one that you started here.

Your LCD is $xy^2z$ (all the different factors of the denominator with the highest exponent)

Multiply that LCD times each numerator and simplify. Apply the results over the LCD of $xy^2z$

$\frac{13}{xy^2}-\frac{6}{yz}$

$\frac{xy^2z(\frac{13}{xy^2})-xy^2z(\frac{6}{yz})}{xy^2z}$

Now simplify the numerator:

$\frac{13z-6xy}{xy^2z}$
• August 15th 2008, 09:14 AM
cmf0106
Thanks Masters! Your advice makes it very clear. I assume this will always work, and I can continue to build on this method even with more advance problems. I cannot thank you enough the concept in the book makes next to no sense to me. Thank you!!!!
• August 15th 2008, 09:16 AM
allyourbass2212
Masters, would you please show your work on how to simplify this problem?Thanks.
$\frac{xy^2z(\frac{13}{xy^2})-xy^2z(\frac{6}{yz})}{xy^2z}
$
• August 15th 2008, 09:29 AM
o_O
$\frac{{\color{red}x}{\color{blue}y^2}z(\frac{13}{{ \color{red}x}{\color{blue}y^2}}) \: - \: x{\color{blue}y^{2}}{\color{magenta}z}(\frac{6}{{\ color{blue}y}{\color{magenta}z}})}{xy^2z}$

Notice that the variables colour-coded can be cancelled out (basically, dividing a number by itself gives you 1 which you don't need to write out).

When dividing powers in the case of ${\color{blue}y^2}$, remember that: $\frac{y^{n}}{y^{m}} = y^{n-m}$. Here, n = 2 and m = 1.
• August 15th 2008, 09:34 AM
masters
Quote:

Originally Posted by allyourbass2212
Masters, would you please show your work on how to simplify this problem?Thanks.
$\frac{xy^2z(\frac{13}{xy^2})-xy^2z(\frac{6}{yz})}{xy^2z}$

$\frac{xy^2z(\frac{13}{xy^2})-xy^2z(\frac{6}{yz})}{xy^2z}= \frac{{\frac{13xy^2z}{xy^2}} - \frac{6xy^2z}{yz}}{xy^2z}$

In the numerator of the complex fraction above, you can simplify by dividing out the variables resulting in:

$\frac{13z-6xy}{xy^2z}$
• August 15th 2008, 09:36 AM
allyourbass2212
This is going to be a real stupid question but in the final answer
$\frac{13}{xy^2}-\frac{6}{yz} =
\frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z}
=\frac{13z-6xy}{xy^2z}
$

Why does it not simply even further? IE There is a single x & y variable in the numerator and denominator, how come those cannot be canceled out and have the answer read $\frac{13-6y}{y^2}$?
• August 15th 2008, 09:44 AM
masters
Quote:

Originally Posted by allyourbass2212
This is going to be a real stupid question but in the final answer
$\frac{13}{xy^2}-\frac{6}{yz} =
\frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z}
=\frac{13z-6xy}{xy^2z}
$

Why does it not simply even further? IE There is a single x & y variable in the numerator and denominator, how come those cannot be canceled out and have the answer read $\frac{13-6y}{y^2}$?

You can only "cancel" across factors, not terms that have been added or subtracted. For instance,

$\frac{xy^2z}{x^2y}=\frac{yz}{x}$ with numerator and denominator being all factors (no + or - signs)
• August 15th 2008, 10:25 AM
allyourbass2212
So you can only "cancel" out that way whenever the numerator and denominator numbers are connected by multiplication, if there is a + or - in the numerator you cannot cancel out that way.

$\frac{13z-6xy}{xy^2z}
$

So in this instance since there is a '-' in the numerator this is as simplified as its going to get.

• January 30th 2009, 11:19 PM
princess_21
Quote:

Originally Posted by cmf0106
I need some help understanding whats going on here.

Description:
When adding fractions with variables in one or more denominators, the LCD will have each variable to its highest power as a factor.
$\frac{13}{xy^2}-\frac{6}{yz} = \frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z} =\frac{13z-6xy}{xy^2z}$

Can someone please help me through this process, I understand picking out the highest exponent of each variable but after that its unclear. Also how do you know to arrange the expression? For instance $\frac{13}{xy^2}*\frac{z}{z}$ How would I know that the $\frac{z}{z}$ gets matched up with the $\frac{13}{xy^2}$ and not on the other end of the expression with the $\frac{6xy}{xy^2z}$?

Many Thanks

you should follow the PEMDAS rule
Parenthesis, Exponent, Multiplication, Division, Addition then Subtraction
:)