- Identify the common denominator as \((x-4)(x+4)\) by factoring \(x^2-16\).
- Rewrite the second term as \(\frac{3(x+4)}{(x-4)(x+4)}\) to have the same denominator.
- Combine the numerators to get \(x - 3(x+4) = -2x - 12\).
- Simplify the final fraction to \(\frac{-2(x+6)}{(x-4)(x+4)}\).
- The correct choice is \(\frac{-2(x+6)}{(x+4)(x-4)}\), so \(\boxed{\frac{-2(x+6)}{(x+4)(x-4)}}\).
Explanation
- Analyze the expression and find a common denominator
We are asked to simplify the expression \(\frac{x}{x^2-16} - \frac{3}{x-4}\). To subtract these two rational expressions, we first need to identify a common denominator. The denominator of the first term is \(x^2 - 16\), which can be factored as \((x-4)(x+4)\) because it is a difference of squares. The denominator of the second term is \((x-4)\). Thus, the common denominator for both terms is \((x-4)(x+4)\).
- Rewrite the fractions with the common denominator
Now, we rewrite each fraction with the common denominator \((x-4)(x+4)\). The first fraction is already in the desired form:
\[\frac{x}{x^2-16} = \frac{x}{(x-4)(x+4)}.\]
For the second fraction, we multiply the numerator and the denominator by \((x+4)\):
\[\frac{3}{x-4} = \frac{3(x+4)}{(x-4)(x+4)}.\]
- Perform the subtraction and simplify the numerator
Now that the fractions have the same denominator, we can subtract them by combining their numerators:
\[\frac{x}{(x-4)(x+4)} - \frac{3(x+4)}{(x-4)(x+4)} = \frac{x - 3(x+4)}{(x-4)(x+4)}.\]
Next, we expand the expression in the numerator:
\[x - 3(x+4) = x - 3x - 12 = -2x - 12.\]
Finally, we can factor out \(-2\) from the numerator:
\[-2x - 12 = -2(x+6).\]
Thus, the expression simplifies to \(\frac{-2(x+6)}{(x-4)(x+4)}\).
- Conclusion
Comparing our result \(\frac{-2(x+6)}{(x-4)(x+4)}\) with the given options, we see that it matches the second option. Therefore, the difference is \(\frac{-2(x+6)}{(x+4)(x-4)}\).
Examples
Simplifying rational expressions is essential in various fields such as electrical engineering when combining resistors in parallel or analyzing signal circuits. For example, if two components in a circuit have impedances described by rational functions, finding the total impedance often requires adding or subtracting these fractions, much like we simplified the expression above to determine the combined behavior of the system.