Once you factor $k^2 - 9$, what denominator could all three fractions share? How can you adjust the second and third fractions so they have this same denominator?

After rewriting all fractions with the same denominator, put them over a single denominator and carefully expand and combine like terms in the numerator. Watch the minus sign in front of $\dfrac{3}{k-3}$.

In Desmos, type the original expression as a function, for example: f(k) = 2k/(k^2 - 9) - 3/(k - 3) + (k + 5)/(k + 3).

Define each option as a new function, such as A(k) = (k^2 - k - 24)/(k^2 - 9), B(k) = (k^2 + k - 24)/(k^2 + 9), C(k) = (k^2 + k - 24)/(k^2 - 6k + 9), and D(k) = (k^2 + k - 24)/(k^2 - 9).

Use a table (click the gear icon next to one function and add a table) to evaluate f(k) and each of A(k), B(k), C(k), and D(k) at several values like $k = 0, 1, 2, 4$ (avoid $k = \pm 3$, where denominators are zero). Look for which option’s values always match $f(k)$.

On the graph, look for the option whose graph lies exactly on top of the graph of f(k) (except at excluded points where there are holes or vertical asymptotes). That function represents the equivalent expression.

Notice that the first denominator is $k^2 - 9$, which is a difference of squares:

So the common denominator for all three fractions can be $(k - 3)(k + 3)$ (which is the same as $k^2 - 9$).

Rewrite each term so it has denominator $(k - 3)(k + 3)$:

• First term: $\dfrac{2k}{k^2 - 9}$ already has denominator $(k - 3)(k + 3)$.
• Second term: $-\dfrac{3}{k - 3}$
  • Multiply top and bottom by $(k + 3)$ to get
  • $-\dfrac{3(k + 3)}{(k - 3)(k + 3)}$.
• Third term: $\dfrac{k + 5}{k + 3}$
  • Multiply top and bottom by $(k - 3)$ to get
  • $\dfrac{(k + 5)(k - 3)}{(k - 3)(k + 3)}$.

Now the expression is

Since all three fractions have the same denominator, combine them into one fraction by adding/subtracting the numerators:

Now expand and simplify the numerator:

• Expand $-3(k + 3) = -3k - 9$.
• Expand $(k + 5)(k - 3) = k^2 + 2k - 15$.

So the numerator becomes

Combine like terms:

• $2k - 3k + 2k = k$.
• $-9 - 15 = -24$.

So the numerator simplifies to $k^2 + k - 24$. The denominator is still $(k - 3)(k + 3) = k^2 - 9$.

The whole expression simplifies to

This matches answer choice D, so $\dfrac{k^{2}+k-24}{k^{2}-9}$ is the equivalent expression.