# Subtracting Fractions

• Mar 30th 2013, 05:30 PM
dsDoan
Subtracting Fractions
"Find the difference and simplify, if necessary."

If you can make out my work and identify my error, that would be great. The initial problem is in the top right, in Arial font. My work begins at the top left, with the equation rewritten, and continues down and to the right. If you cannot make out my work, the correct steps would be great, as well.

Also, how are symbols for equations entered directly in the post text? I don't see any format options for them.
• Mar 30th 2013, 06:44 PM
Soroban
Re: Subtracting Fractions
Hello, dsDoan!

Quote:

$\displaystyle \frac{-7x - 49}{x^2 + 2x-35} - \frac{x+7}{5-x}$

We have: .$\displaystyle \frac{-7x-49}{(x-5)(x+7)} - \frac{x+7}{{\color{red}-}(x-5)} \;=\;\frac{-7x-49}{(x-5)(x+7)} \;{\color{red}+}\; \frac{x+7}{x-5}$

. . . . . .$\displaystyle =\;\;\frac{-7x-49}{(x-5)(x+78)} + \frac{x+7}{x-5}\cdot {\color{blue}\frac{x+7}{x+7}} \;\;=\;\;\frac{-7x-49}{(x-5)(x+7)} + \frac{(x+7)^2}{(x-5)(x+7)}$

. . . . . .$\displaystyle =\;\;\frac{-7x-49 + (x+7)^2}{(x-5)(x+7)} \;\;=\;\; \frac{-7x-49 + x^2 + 14x + 49}{(x-5)(x+7)}$

. . . . . .$\displaystyle =\;\;\frac{x^2+7x}{(x-5)(x+7)} \;\;=\;\;\frac{x(x+7)}{(x-5)(x+7)} \;\;=\;\;\frac{x}{x-5}$
• Mar 30th 2013, 08:03 PM
dsDoan
Re: Subtracting Fractions
My mistake was with a negative sign from the start, so I'd like to get that issue straightened out:

$\displaystyle -\frac{x+7}{5-x}\;=\; +\frac{x+7}{x-5}$

Going from step one to step two, a negative one has been distributed to the denominator. Why does this make the fraction positive?
• Mar 30th 2013, 10:02 PM
ibdutt
Re: Subtracting Fractions
we can also do it as shown in the attachment.Attachment 27744
• Mar 31st 2013, 06:49 AM
dsDoan
Re: Subtracting Fractions
Quote:

Originally Posted by ibdutt
we can also do it as shown in the attachment.Attachment 27744

This is the method I originally used when starting this section, but cancelling early in the equation doesn't always lead to two common denominators so I discontinued this approach. If I realize, early on, that cancelling early will result in common denominators it would save time, so I'll keep this in mind.

I would still like to clear up my issue with the negative. It seems like such a trivial issue that will be an ongoing problem if I don't get it sorted now.
• Mar 31st 2013, 08:46 PM
ibdutt
Re: Subtracting Fractions
There are two ways to look at it. One is that we multiply and divide by negative one OR alternatively be tale negative one common. In both cases we will have negative into negative and that is positive.
• Apr 2nd 2013, 03:31 PM
dsDoan
Re: Subtracting Fractions
Quote:

Originally Posted by ibdutt
There are two ways to look at it. One is that we multiply and divide by negative one OR alternatively be tale negative one common. In both cases we will have negative into negative and that is positive.

Can you explain this by applying it to the example I posted in post #3?