Question

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

Explanation:

Step1: Factor the denominator

Factor \(d^2 - 2d - 8\): \(d^2 - 2d - 8=(d - 4)(d + 2)\)
The equation becomes \(\frac{-3d}{(d - 4)(d + 2)}+\frac{3}{d - 4}=\frac{-2}{d + 2}\)

Step2: Find the common denominator

The common denominator is \((d - 4)(d + 2)\). Multiply each term by the common denominator to eliminate fractions:
\(-3d+3(d + 2)=-2(d - 4)\)

Step3: Simplify both sides

Expand the left - hand side: \(-3d+3d + 6=-2(d - 4)\)
Simplify the left - hand side: \(6=-2d + 8\)

Step4: Solve for d

Subtract 8 from both sides: \(6-8=-2d\)
\(-2=-2d\)
Divide both sides by - 2: \(d = 1\)

We need to check for extraneous solutions. The original equation has restrictions \(d
eq4\) and \(d
eq - 2\) (because these values make the denominators zero). When \(d = 1\), the denominators are non - zero.

Answer:

\(d = 1\)