Why can't I shorten this fraction?

[Vincent_M05]

If you break this question down to the point where you will have:


Why can't you shorten the fraction like this:

Since I thought that you could do it this way?

Thanks for every help.

 

[BigBeachBanana]

To your answer your question, [math]\frac{9+2}{3}=\frac{11}{3}[/math]However, by your logic
[math]\frac{\sout{9}+2}{\sout{3}} = 5[/math]which is wrong
You can only cancel when both your numerators and denominators are composed of multiplications only.
[math]\frac{9*2}{3} =\frac{\sout{9}*2}{\sout{3}}=6[/math]

 

[Vincent_M05]

First, I don't think you added the fractions correctly. Would you please show the intermediate steps?
To your answer your question, [math]\frac{9+2}{3}=\frac{11}{3}[/math]However,
[math]\frac{\sout{9}+2}{\sout{3}} \neq 5[/math]You can only cancel when both your numerators and denominators are composed of multiplications only.
[math]\frac{9*2}{3} =\frac{\sout{9}*2}{\sout{3}}=6[/math]

Of course

 

[Steven G]

To add to what BBB wrote consider this argument.

Note that in \(\displaystyle \dfrac{9}{3}+\dfrac{2}{3}\) that BOTH the 9 and the 2 are being divided by 3.

Now we know that \(\displaystyle \dfrac{9}{3}+\dfrac{2}{3}=\dfrac{9+2}{3} \neq\dfrac{9}{3} + 2 = 5\) since both the 9 and the 2 must be divided by 3

 

[Vincent_M05]

To add to what BBB wrote consider this argument.

Note that in \(\displaystyle \dfrac{9}{3}+\dfrac{2}{3}\) that BOTH the 9 and the 2 are being divided by 3.

Now we know that \(\displaystyle \dfrac{9}{3}+\dfrac{2}{3}=\dfrac{9+2}{3} \neq\dfrac{9}{3} + 2 = 5\) since both the 9 and the 2 must be divided by 3

yep, big thanks to both of you. I followed your example and understood the problem, thanks a lot.

 

[Steven G]

Of course
View attachment 30375

You should distribute that 3 in your final answer since *maybe* you could then factor the numerator and have a factor of 2, (x-1) and/or x+1. Does this make sense?