Simplifying (k + 7/k^2 + 6K + 9) + (k - 5/k^2 - 5k - 24)

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Discussion Overview

The discussion revolves around simplifying the expression involving two rational functions: (k + 7)/(k^2 + 6k + 9) + (k - 5)/(k^2 - 5k - 24). Participants explore the process of factoring the denominators and finding a common denominator to combine the fractions.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Some participants clarify the expression to be simplified as $$\dfrac{k + 7}{k^2 + 6k + 9} + \dfrac{k - 5}{k^2 - 5k - 24}$$.
  • There is a suggestion that both denominators should factor nicely, although the relevance of this to finding the sum is questioned.
  • One participant states that after factoring, they arrive at $$\frac{k+7}{(k+3)^2}+\frac{k-5}{(k-8)(k+3)}$$.
  • Another participant confirms the common denominator (LCD) as (k - 8)(k + 3)^2 and discusses the process of multiplying fractions by their missing factors to achieve a common denominator.
  • There is a proposal to combine the two terms into a single fraction and to simplify the numerator by expanding and combining like terms.

Areas of Agreement / Disagreement

Participants generally agree on the need to factor the denominators and find a common denominator, but there is no consensus on the final simplification steps or the correctness of the factored forms presented.

Contextual Notes

Some assumptions about the factorability of the denominators and the steps involved in simplification remain unresolved, as participants have not fully explored the implications of their factorizations.

elsacozine
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(k + 7/k^2 + 6K + 9) + (k - 5/k^2 - 5k - 24)
 
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Hi elsacozine and welcome to MHB! :D

Do you mean "Find the sum:

$$\dfrac{k + 7}{k^2 + 6k + 9} + \dfrac{k - 5}{k^2 - 5k - 24}$$

?
 
Both denominators should factor nicely.
Although true, is it really relevant to finding the sum?
 
suluclac said:
Both denominators should factor nicely.
Although true, is it really relevant to finding the sum?
Yes . . . we need to find a common denominator.

 
greg1313 said:
Hi elsacozine and welcome to MHB! :D

Do you mean "Find the sum:

$$\dfrac{k + 7}{k^2 + 6k + 9} + \dfrac{k - 5}{k^2 - 5k - 24}$$

?

Yes; I meant that. Sorry I didn't know how to write it like that.
 
elsacozine said:
Yes; I meant that. Sorry I didn't know how to write it like that.

Okay, so what you want to do is first factor the two denominators (if possible), so that you can determine the LCD. What do you get when factoring?
 
MarkFL said:
Okay, so what you want to do is first factor the two denominators (if possible), so that you can determine the LCD. What do you get when factoring?

when i factor it i get:
(k + 7)/(k + 3)(k + 3) + (k - 5)/(k - 8)(k + 3)
 
elsacozine said:
when i factor it i get:
(k + 7)/(k + 3)(k + 3) + (k - 5)/(k - 8)(k + 3)

Yes, and we can write this as:

$$\frac{k+7}{(k+3)^2}+\frac{k-5}{(k-8)(k+3)}$$

So, what is our LCD?
 
MarkFL said:
Yes, and we can write this as:

$$\frac{k+7}{(k+3)^2}+\frac{k-5}{(k-8)(k+3)}$$

So, what is our LCD?

the LCD is (k - 8)(k + 3)^2...
so would you multiply the fractions by the their missing factors. then the denominators should be the same (LCD), then you could add the two fractions because they have common denominators?
 
  • #10
elsacozine said:
the LCD is (k - 8)(k + 3)^2...
so would you multiply the fractions by the their missing factors. then the denominators should be the same (LCD), then you could add the two fractions because they have common denominators?

Yes, although what you actually want to do is multiply each term by 1 in the form of the missing factor in that term's denominator divided by itself so that all terms have the same denominator:

$$\frac{k+7}{(k+3)^2}\cdot\frac{k-8}{k-8}+\frac{k-5}{(k-8)(k+3)}\cdot\frac{k+3}{k+3}$$

And so now you can combine the two terms:

$$\frac{(k+7)(k-8)+(k-5)(k+3)}{(k+3)^2(k-8)}$$

At this point you would likely want to see if you can simplify the numerator by expanding, combining like terms and attempting to factor. :)
 

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