- Factor the denominator of the first term: \(2x^2 - 4x = 2x(x - 2)\).
- Identify the common denominator as \(2x(x - 2)\).
- Rewrite the second term to share the common denominator: \(\frac{4(x - 2)}{2x(x - 2)}\).
- Combine and simplify: \(\frac{1 - (4x - 8)}{2x(x - 2)} = \frac{-4x + 9}{2x(x - 2)}\).
- The final answer is \(\boxed{\frac{-4x + 9}{2x(x - 2)}}\).
Explanation
- Factoring the Denominator
The goal is to simplify the rational expression \(\frac{1}{2x^2 - 4x} - \frac{2}{x}\). To subtract two rational expressions, we must first ensure they have a common denominator. The first step is to factor the denominator of the first expression. The denominator is \(2x^2 - 4x\), and we can factor out the greatest common factor, \(2x\), which gives \(2x(x - 2)\). Thus, the expression becomes \(\frac{1}{2x(x - 2)} - \frac{2}{x}\).
- Finding a Common Denominator
Now that we have factored the denominator, we can clearly see the common denominator is \(2x(x - 2)\). The first term \(\frac{1}{2x(x - 2)}\) already has this denominator. To express the second term \(\frac{2}{x}\) with the same denominator, we multiply its numerator and denominator by \(2(x - 2)\):
\[\frac{2}{x} \cdot \frac{2(x - 2)}{2(x - 2)} = \frac{4(x - 2)}{2x(x - 2)}\]
Now both terms have the same denominator:
\[\frac{1}{2x(x - 2)} - \frac{4(x - 2)}{2x(x - 2)}\]
- Combining and Simplifying the Numerator
With a common denominator, we can combine the numerators:
\[\frac{1 - 4(x - 2)}{2x(x - 2)}\]
Next, distribute the \(-4\) in the numerator:
\[1 - 4x + 8\]
Combine the constant terms:
\[9 - 4x\]
Thus, the simplified expression is:
\[\frac{-4x + 9}{2x(x - 2)}\]
This matches the second option provided in the problem.
Examples
Rational expressions are frequently used in physics and engineering to model systems where variables are inversely related. For example, if you are designing a circuit where resistors are connected in parallel, the total resistance is calculated using the reciprocal of the sum of individual resistances, which often involves simplifying rational expressions like the one solved here. Similarly, in fields like fluid dynamics or thermodynamics, understanding how to simplify these algebraic forms is crucial for balancing equations that describe heat exchange or fluid flow rates.