# Thread: Combine these into a single equivalent fraction;

1. ## Combine these into a single equivalent fraction;

Combine these into a single equivalent fraction;

3/K - 4 + K + 4/K^2 - 16

I factored k^2 - 16 so that its (k - 4) (k + 4)

The k + 4 cancel out as like expressions

and I end up with k - 4 as the denominator and 3 as the numerator such as..

3/k - 4

Is this correct?

2. Here is what you have written:

$\frac{3}{K} - 4 + k + \frac{4}{k^{2}} - 16$

I assume you mean $\frac{3}{k-4}+\frac{k+4}{k^{2}-16}$

I am only guessing. Without proper grouping symbols, it can mean various things.

3. Originally Posted by galactus
Here is what you have written:

$\frac{3}{K} - 4 + k + \frac{4}{k^{2}} - 16$

I assume you mean $\frac{3}{k-4}+\frac{k+4}{k^{2}-16}$

I am only guessing. Without proper grouping symbols, it can mean various things.
Yes, that is the correct Interpretation .

4. Originally Posted by JeffM
Now that we know that, we can explain why your answer is wrong. It greatly helps to show your work. You will catch many of your own errors if you do.

$\dfrac{3}{k-4}+\dfrac{k+4}{k^2-16} =$

$\dfrac{3}{k-4}+\dfrac{k+4}{(k + 4)(k-4)} =$

$\dfrac{3}{k-4} + \dfrac{1}{k-4} =$

$\dfrac{4}{k-4}.$

How did you get 1 from (k + 4) and then add it to 3 to get 4 for the numerator?

Oh, I see you eliminated (k + 4) in both numerator and denominators and your left with a 1 for the numerator?

5. one restriction should be included:

$k \ne \pm 4$

6. Originally Posted by gijas
3/K - 4 + K + 4/K^2 - 16
Next time make SURE you bracket properly: 3 / (K - 4) + (K + 4) / (K^2 - 16)

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