Question

1. \\(\frac{k + 2}{k^2 - 8k + 15} + \frac{2k + 4}{k^2 - 8k + 15}\\)

Explanation:

Step1: Combine the fractions

Since the denominators are the same ($k^2 - 8k + 15$), we can add the numerators:
$\frac{(k + 2) + (2k + 4)}{k^2 - 8k + 15}$

Step2: Simplify the numerator

Combine like terms in the numerator:
$k + 2 + 2k + 4 = 3k + 6$
So now we have $\frac{3k + 6}{k^2 - 8k + 15}$

Step3: Factor numerator and denominator

Factor the numerator: $3k + 6 = 3(k + 2)$
Factor the denominator: $k^2 - 8k + 15 = (k - 3)(k - 5)$ (by finding two numbers that multiply to 15 and add to -8, which are -3 and -5)
So we get $\frac{3(k + 2)}{(k - 3)(k - 5)}$ (Note: We can check if we can cancel any factors, but $(k + 2)$ doesn't match $(k - 3)$ or $(k - 5)$, so this is the simplified form. Also, we should note the domain restrictions: $k
eq 3$ and $k
eq 5$ to avoid division by zero.)

Answer:

$\frac{3(k + 2)}{(k - 3)(k - 5)}$ (with $k
eq 3, k
eq 5$)