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

[elsacozine]

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

 

[Greg]

Hi elsacozine and welcome to MHB! :D

Do you mean "Find the sum:

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

?

 

[suluclac]

Both denominators should factor nicely.
Although true, is it really relevant to finding the sum?

 

[soroban]

Both denominators should factor nicely.
Although true, is it really relevant to finding the sum?

Yes . . . we need to find a common denominator.

 

[elsacozine]

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.

 

[MarkFL]

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?

 

[elsacozine]

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)

 

[MarkFL]

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?

 

[elsacozine]

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?

 

[MarkFL]

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. :)