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

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SUMMARY

The discussion focuses on simplifying the expression $$\dfrac{k + 7}{k^2 + 6k + 9} + \dfrac{k - 5}{k^2 - 5k - 24}$$ by factoring the denominators and finding a common denominator (LCD). The denominators factor to $(k + 3)^2$ and $(k - 8)(k + 3)$. The LCD is established as $(k - 8)(k + 3)^2$, allowing for the combination of the fractions. The final expression is $$\frac{(k+7)(k-8)+(k-5)(k+3)}{(k+3)^2(k-8)}$$, which can be further simplified by expanding and combining like terms.

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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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