Possible Misunderstanding with Cancelling Algebra: simplify by making into a simple fraction. x+1/x-4 - x-1/(x-4)^2

[mhslin0107]

The objective is to simplify by making into a simple fraction.

x+1/x-4 - x-1/(x-4)^2

when I put this into an algebra math solver so that it can show me the steps so I can improve the next time, it says that after we add the two fractions together (multiply first fraction by x-4 on both top and bottom,) we don't cancel out the bottom and top (x-4)s. But instead multiplies (x-4) by (x+1) on the top. And then simplies from there. I just don't understand why you would simplify (x-4) on the top with (x-4)^2 on the bottom? cant it be simplified to remove the top one and the bottom one is leaved with (x-4)?

 

[blamocur]

What you are describing sounds like bringing two fractions to common denominator -- are you familiar with this technique?
It would be easier to help you if you posted the formulas you get in your transformations, and please do not forget to use parenthesis. I believe you meant to write "(x+1)/(x-4) - (x-1)/(x-4)^2". You could make it even more readable if you used LaTex (KaTex):
[math]\frac{x+1}{x-4}-\frac{x-1}{(x-4)^2}[/math]

 

[rich20509]

As it is already mention above multiply both the numerator and denominator of the first fraction by (x-4) Like given in the image below

 

[lookagain]

As it is already mention above multiply both the numerator and denominator of the first fraction by (x-4) Like given in the image below

@ rich20509 -- You are new here, and you should not have posted an alleged
complete solution to the OP's question from two days ago. The OP did not
respond back with a folow-up post.

Most of your post is wrong.

Second line: The (x + 1) factor and the (x - 4) factor are missing required
grouping symbols. The fraction bar is not sufficiently extended on each end.

Fifth line: The minus sign was not distributed on the (x - 1) factor, and there
is an unmatched parenthesis at the end.

Sixth line: As the result of the error in the fifth line, the expression for the
numerator is wrong here.

Last line: Therefore, the expression for the simplified fraction is also wrong.

 

[The Highlander]

The objective is to simplify by making into a simple fraction.

x+1/x-4 - x-1/(x-4)^2

when I put this into an algebra math solver so that it can show me the steps so I can improve the next time, it says that after we add the two fractions together (multiply first fraction by x-4 on both top and bottom,) we don't cancel out the bottom and top (x-4)s. But instead multiplies (x-4) by (x+1) on the top. And then simplies from there. I just don't understand why you would simplify (x-4) on the top with (x-4)^2 on the bottom? cant it be simplified to remove the top one and the bottom one is leaved with (x-4)?

If, indeed, the original problem was given as:-

\(\displaystyle \frac{x+1}{x-4}-\frac{x-1}{(x-4)^2}\qquad\) (Courtesy of @blamocur)​

then you appear to have understood that the two fractions need to have the same denominator (before you can add or subtract them) and so you need to multiply both top and bottom of the left one by (x - 4). to get:-

[math]\frac{(x+1)\times (x-4)}{(x-4)\times(x-4)}-\frac{x-1}{(x-4)^2}\\ \, \\=\frac{(x+1)(x-4)}{(x-4)^2}-\frac{(x-1)}{(x-4)^2}[/math]
Now that the two fractions have the same denominator you can subtract the right numerator from the left one to get a single fraction, thus...
[math]\frac{(x+1)(x-4)-(x-1)}{(x-4)^2}[/math]
To answer your original question, when it is laid out like this can you now see why you cannot "cancel" the (x - 4) on the top with (x - 4) on the bottom?

The (x - 4) on the top is only multiplying the (x + 1) on the top, it is not multiplying the whole of the numerator, therefore, it is not a factor of both the top and the bottom so you cannot divide both top and bottom by it (without leaving the numerator in a very messy state).

Please tell us you now understand and maybe you can now go on to simplify the numerator (ignoring the rubbish @rich20509 posted) and show us what simple(r) fraction you get at the end?

Hope that helps.

 

[rich20509]

To be honest I was just trying to help but don't you think this is not the way to talk to a newbie as I was not aware of this community.
I am really disappointed with this website: https://www.freemathhelp.com/ forum and regret to join it.

As you are an old member it doesn't mean you have all the rights to disrespect any newbie.
@The Highlander

 

[mario99]

To be honest I was just trying to help but don't you think this is not the way to talk to a newbie as I was not aware of this community.
I am really disappointed with this website: https://www.freemathhelp.com/ forum and regret to join it.

As you are an old member it doesn't mean you have all the rights to disrespect any newbie.
@The Highlander

I respect you rich. And I think that the help that you have given was wonderful. It was only given in the wrong time. Everyone makes mistakes and learn from their mistakes.

 

[The Highlander]

I respect you rich. And I think that the help that you have given was wonderful. It was only given in the wrong time. Everyone makes mistakes and learn from their mistakes.

I'm afraid the "help" given by @rich20509 wasn't just given at the "wrong time" and neither was it "wonderful" because it was wrong!

@lookagain chided @rich20509 for posting a "complete solution" just 2 days after the OP and before the author had made any response. There is nothing in the Forum Rules that specifically prohibits posting solutions (complete or otherwise) at any time but there has been extensive discussion on the subject (here, qv) and the consensus reached (among the Forum's regular contributors) was that at least a 4 day gap should be left between an original post and any solution being posted.

Trusted (long-term) members often get away with ignoring that decision but newer members risk having their post(s) deleted if they transgress (especially, if having only recently joined the site, they appear to be just trying to show (off) how 'good' they are at solving some of the problems posted).

To be honest I was just trying to help but don't you think this is not the way to talk to a newbie as I was not aware of this community.
I am really disappointed with this website: https://www.freemathhelp.com/ forum and regret to join it.

As you are an old member it doesn't mean you have all the rights to disrespect any newbie.
@The Highlander

However, I did not describe @rich20509's post as "rubbish" because it was posted too 'early' but because, not only was it poor Algebra, it didn't even address the OP's problem!

The OP expressed a difficulty in understanding why s/he couldn't cancel out the (x - 4) term from both the numerator and the denominator; my post (#5, qv) attempted to explain that (without providing the final answer to the original problem).

Yes, we all make mistakes but when we do (especially if they are as glaring as that made by @rich20509) then we must be prepared to accept that they may be criticised and, depending upon the severity of the error, severely so.

Your attempt to "help" did not only ignore the OP's actual problem but, because it was Algebraically incorrect, it could have caused further confusion!

My characterisation of your post was perfectly warranted and your suggestion that you have been disrespected is not; you need to be sure (check your work before posting) that what you are writing is correct if you wish to avoid any adverse comments on it.

If you are unhappy with how the Forum operates then nobody is forcing you to revisit it.