Question

what is the solution to the equation \\(\frac{-3d}{d^2 - 2d - 8} + \frac{3}{d - 4} = \frac{-2}{d + 2}\\)?\
\\(\circ\\) \\(d = -4\\) and \\(d = 2\\)\
\\(\circ\\) \\(d = -2\\) and \\(d = 4\\)\
\\(\circ\\) \\(d = 1\\)\
\\(\circ\\) \\(d = 2\\)

Explanation:

Step1: Factor denominator

Factor $d^2-2d-8$ into $(d-4)(d+2)$.

Step2: Eliminate denominators

Multiply all terms by $(d-4)(d+2)$:
$$-3d + 3(d+2) = -2(d-4)$$

Step3: Expand all terms

$$-3d + 3d + 6 = -2d + 8$$

Step4: Simplify and solve for d

Combine like terms:
$$6 = -2d + 8$$
Rearrange to isolate $d$:
$$-2d = 6 - 8 = -2$$
$$d = \frac{-2}{-2} = 1$$

Step5: Check excluded values

Denominators cannot be zero, so $d
eq 4$ and $d
eq -2$. $d=1$ is valid.

Answer:

$d=1$ (Option: $d=1$)