# Math Help - Adding/Subracting Fractions with Variables

1. ## Adding/Subracting Fractions with Variables

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

2. 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:

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

3. 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!!!!

4. 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}
$

5. $\frac{{\color{red}x}{\color{blue}y^2}z(\frac{13}{{ \color{red}x}{\color{blue}y^2}}) \: - \: x{\color{blue}y^{2}}{\color{magenta}z}(\frac{6}{{\ color{blue}y}{\color{magenta}z}})}{xy^2z}$

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.

6. 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:

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

7. 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 $\frac{13-6y}{y^2}$?

8. 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)

9. 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.

10. 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