- Identify the common denominator \(2(x-4)\) for the two fractions.
- Rewrite \(\frac{-4}{x-4}\) as \(\frac{-8}{2(x-4)}\).
- Combine the numerators to obtain \(\frac{x^2 - 2x - 8}{2(x-4)}\).
- Factor the numerator as \((x-4)(x+2)\) and simplify to reach the final answer of \(\boxed{\frac{x+2}{2}}\).
Explanation
- Analysis of the Expression
The given expression is \(\frac{-4}{x-4} + \frac{x^2 - 2x}{2x - 8}\). Our objective is to combine these rational expressions into a single, simplified form. First, we identify the denominators: the first is \(x-4\) and the second is \(2x-8\). Notice that the second denominator can be factored as \(2(x-4)\). Thus, the least common denominator for both expressions is \(2(x-4)\). We must note that for the expression to be defined, \(x
eq 4\).
- Combining the Fractions
To add the fractions, we need to express the first term with the common denominator \(2(x-4)\). We do this by multiplying the numerator and the denominator of the first term by 2:
\[\frac{-4}{x-4} = \frac{-4 \times 2}{(x-4) \times 2} = \frac{-8}{2(x-4)}\]
Now, we can combine the two fractions:
\[\frac{-8}{2(x-4)} + \frac{x^2 - 2x}{2(x-4)} = \frac{x^2 - 2x - 8}{2(x-4)}\]
- Factoring and Simplifying the Expression
Now we focus on the numerator \(x^2 - 2x - 8\). We look for factors of \(-8\) that add up to \(-2\). The numbers \(-4\) and \(2\) satisfy these conditions since \((-4) \times 2 = -8\) and \(-4 + 2 = -2\). Therefore, we can factor the numerator as follows:
\[x^2 - 2x - 8 = (x-4)(x+2)\]
Substituting this back into our expression, we get:
\[\frac{(x-4)(x+2)}{2(x-4)}\]
- Final Simplification
Since \(x
eq 4\), the term \((x-4)\) is non-zero, and we can cancel the common factor \((x-4)\) from the numerator and the denominator:
\[\frac{(x-4)(x+2)}{2(x-4)} = \frac{x+2}{2}\]
Comparing this result with the given choices, the simplified expression is \(\frac{x+2}{2}\).
Examples
Rational expressions often arise in physics when calculating the effective resistance of circuits or in economics when determining average costs. For example, if you are analyzing the unit production cost of a product, you might have expressions involving a fixed setup cost divided by a variable number of units, plus a variable cost component. Simplifying these expressions helps in quickly understanding the relationship between the number of units produced and the resulting total cost, allowing for better decision-making in manufacturing.