Question

what is the solution to the equation \\(\frac{m}{m+4} + \frac{4}{4-m} = \frac{m^2}{m^2 - 16}\\)?
\\(m = -4\\) \\(m = 2\\) \\(m = 4\\)
\\(m = -2\\)

Explanation:

Step1: Factor Denominators

Notice that \(m^2 - 16=(m + 4)(m - 4)\) and \(4 - m=-(m - 4)\). So the equation \(\frac{m}{m + 4}+\frac{4}{4 - m}=\frac{m^2}{m^2 - 16}\) can be rewritten as \(\frac{m}{m + 4}-\frac{4}{m - 4}=\frac{m^2}{(m + 4)(m - 4)}\).

Step2: Multiply by LCD

The least common denominator (LCD) of the fractions is \((m + 4)(m - 4)\). Multiply each term by \((m + 4)(m - 4)\) to eliminate the denominators:

$$
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Step3: Simplify and Solve

Simplify the left - hand side: \(m^2-8m - 16=m^2\). Subtract \(m^2\) from both sides: \(-8m-16 = 0\). Add 16 to both sides: \(-8m=16\). Divide both sides by \(-8\): \(m=-2\).

We also need to check for extraneous solutions. The original equation is undefined when \(m = 4\) or \(m=-4\) (since these values make the denominators zero). When \(m=-2\), the denominators \(m + 4=-2 + 4 = 2\), \(4 - m=4+2 = 6\), and \(m^2-16=4 - 16=-12\) are all non - zero. So \(m=-2\) is a valid solution.

Answer:

\(m=-2\) (the option with \(m = - 2\))