Determing the simplified form of a difference quotient

[Grimsuke]

So I have to simplify this equation... (Sorry it's orientation weirdly, I don't know how to fix that) An online calculator tells me that the answer is but it seems way too complex to be the answer. And why would the denominator be that? It's minus, not multiplication. Where did I go wrong?

 

[Dr.Peterson]

For those of us whose necks don't turn easily anymore, here is the image:



The denominator in the answer is the product of the two denominators, because they have added the fraction using that common denominator.

Give it a try!

 

[Otis]

… why would the denominator be that? It's [subtraction], not multiplication …

Hi Grimsuke. We need a common denominator, when subtracting a fraction. Multiplying denominators gives us a common one.

The concept is the same as when we do this with actual numbers (instead of symbolic numbers).

\[\frac{4}{5} - \frac{3}{7}\]

\[\frac{4(7)}{5(7)} - \frac{3(5)}{7(5)}\]

\[\frac{13}{5(7)}\]

?

 

[Grimsuke]

For those of us whose necks don't turn easily anymore, here is the image:

View attachment 17307

The denominator in the answer is the product of the two denominators, because they have added the fraction using that common denominator.

Give it a try!

So the answer would be that first picture on the original post.. But where does the negative in front come from? What am I missing?

 

[Grimsuke]

Hi Grimsuke. We need a common denominator, when subtracting a fraction. Multiplying denominators gives us a common one.

The concept is the same as when we do this with actual numbers (instead of symbolic numbers).

\[\frac{4}{5} - \frac{3}{7}\]

\[\frac{4(7)}{5(7)} - \frac{3(5)}{7(5)}\]

\[\frac{13}{5(7)}\]

?

Thank you! I definitely needed a refresher in that! It's been a while since I've been in a math course.

 

[Steven G]

The negative in front is because the numerator is negative. The neg sign could have been put in the numerator instead of in the front.

For example,
\(\displaystyle \frac{3}{7} - \frac{4}{5}\)

\(\displaystyle =\frac{3(5)}{7(5)} - \frac{4(7)}{5(7)}\)

\(\displaystyle =\frac{-13}{5(7)}\)

\(\displaystyle =\frac{-13}{35} \ or -\frac{13}{35}\)

 

[Otis]

Thank you! I definitely needed a refresher …

You're welcome, but did you see the multiplications in the numerators? You need to do the same.

\[\frac{2}{2x + 2h - 1} - \frac{2}{2x - 1}\]

The common denominator is (2x-1)(2x+2h-1). We need to convert each ratio above (to get the common denominator) before subtracting the second one from the first.

The second ratio has (2x-1) in the denominator, so we multiply both the numerator and denominator by (2x+2h-1).

\[\frac{2}{2x - 1} · \frac{2x + 2h - 1}{2x + 2h - 1}\]

Likewise, the first ratio has (2x+2h-1) in the denominator, so we multiply both the numerator and denominator by (2x-1).

Can you write it all out? The subtraction takes place in the numerators. Use the distributive property, to carry out the multiplications in the numerators; then combine like-terms. (Don't forget that all the terms in the second numerator are being subtracted, so be mindful of the signs.) You ought to get -4h.

\(\;\)

 

[Steven G]

You're welcome, but did you see the multiplications in the numerators? You need to do the same.

\[\frac{2}{2x - 1} - \frac{2}{2x + 2h - 1}\]

The common denominator is (2x-1)(2x+2h-1). We need to convert each ratio above (to get the common denominator) before subtracting the second one from the first.

The first ratio has (2x-1) in the denominator, so we multiply both the numerator and denominator by (2x+2h-1).

\[\frac{2}{2x - 1} · \frac{2x + 2h - 1}{2x + 2h - 1}\]

Likewise, the second ratio has (2x+2h-1) in the denominator, so we multiply both the numerator and denominator by (2x-1).

Can you write it all out? The subtraction takes place in the numerators. Use the distributive property, to carry out the multiplications in the numerators; then combine like-terms. (Don't forget that all the terms in the second numerator are being subtracted, so be mindful of the signs.) You ought to get -4h.

\(\;\)

To OP, Otis wrote the subtraction in the wrong order. Please keep that in mind.

 

[Otis]

… Otis wrote the subtraction in the wrong order …

Thanks, Jomo. Dang. I'll fix that.

?