subtracting rational expressions

[silverdragon316]

The problem is:

x - 1
x^2 -x -6 x^2 -2x -8


then

u] x [/u] times x-4 - 1 times x-3
x^2 -x -6 times x-4 x^2 -2x -8 times x-3

I stuck here if I am even doing it right in the first place. Help :?

 

[stapel]

After this amount of time (and this number of posts), you should have noticed that your multi-line, space-delimited "formatting" has never yet worked, and that it makes your posts almost entirely unintelligible. You are asking volunteers to help you; please make the effort to speak clearly.

Thank you.

The problem is:...

What were the instructions? Are you supposed to "simplify", "find the domain", "state the asymptotes", or something else? (The instructions are an integral and crucial part of the exercise, and should always be included. Please do so in the future. Thank you.)

x - 1
x^2 -x -6 x^2 -2x -8

Does the above mean the following?

. . . . .[ x / (x^2 - x - 6) ] - [ 1 / (x^2 - 2x - 8) ]

...which may also be formatted in LaTeX as:

. . . . .\(\displaystyle \L \frac{x}{x^2\, -\, x\, -\, 6}\, -\, \frac{1}{x^2\, -\, 2x\, -\, 8}\)

Or did you mean something else?

Please reply with correction or confirmation.

Note: I couldn't guess what you mean in your "work", so I cannot comment on that. When you reply, please include a clear listing of what you have tried so far.

Thank you.

Eliz.

 

[soroban]

Hello, silverdragon316!

Your work is fine . . .

\(\displaystyle \L\frac{x}{x^2\,-\,x\,-\,6}\:-\:\frac{1}{x^2\,-\,2x\,-\,8}\)

then: \(\displaystyle \L\:\frac{x}{(x\,+\,2)(x\,-\,3)}\,\cdot\frac{x\,-\,4}{x\,-\,4}\:-\:\frac{1}{(x\,+\,2)(x\,-\,4)}\.\cdot\,\frac{x\,-\,3}{x\,-\,3}\)

So we have: \(\displaystyle \L\:\underbrace{\frac{x(x\,-\,4)}{(x\,+\,2)(x\,-\,3)(x\,-\,4)}}\:-\:\underbrace{\frac{x\,-\,3}{(x\,+\,2)(x\,-\,3)(x\,-\,4)}}\)
. . . . . . . . . . . . . . . . . . .\(\displaystyle \uparrow\;\,\text{common denominator}\;\,\uparrow\)


Then: \(\displaystyle \L\:\frac{x(x\,-\,4)\,-\,(x\,-\,3)}{(x\,+\,2)(x\,-\,3)(x\,-\,4)} \;=\;\frac{x^2\,-\,4x\,-\,x\,+\,3}{(x\,+\,2)(x\,-\,3)(x\,-\,4)}\)


Therefore: \(\displaystyle \L\:\frac{x^2\,-\,5x\,+\,3}{(x\,+\,2)(x\,-\,3)(x\,-\,4)}\)

 

[silverdragon316]

The problem is:

x - 1
x^2 -x -6 x^2 -2x -8


then

u] x [/u] times x-4 - 1 times x-3
x^2 -x -6 times x-4 x^2 -2x -8 times x-3

I stuck here if I am even doing it right in the first place. Help :?

Thank you for your and I will try my best to make my postings clear. My apologies and thank you for your time.