- Simplify the numerator: \(\frac{3}{x-1} - 4 = \frac{7-4x}{x-1}\).
- Simplify the denominator: \(2 - \frac{2}{x-1} = \frac{2(x-2)}{x-1}\).
- Divide the numerator by the denominator: \(\frac{7-4x}{x-1} \cdot \frac{x-1}{2(x-2)} = \frac{-4x+7}{2(x-2)}\).
- The equivalent expression is \(\boxed{\frac{-4x+7}{2(x-2)}}\).
Explanation
- Analysis and Domain Determination
To simplify the complex fraction \(\frac{\frac{3}{x-1}-4}{2-\frac{2}{x-1}}\), we first identify the domain of the expression. The expression is defined when \(x-1
eq 0\), so \(x
eq 1\), and when the denominator \(2-\frac{2}{x-1}
eq 0\). Solving \(2 - \frac{2}{x-1} = 0\) gives \(2 = \frac{2}{x-1}\), which implies \(x-1 = 1\), or \(x=2\). Thus, the expression is valid for \(x
eq 1\) and \(x
eq 2\).
- Simplifying Numerator and Denominator
Next, we simplify the numerator and the denominator separately.
The numerator is:
\[ \frac{3}{x-1} - 4 = \frac{3 - 4(x-1)}{x-1} = \frac{3 - 4x + 4}{x-1} = \frac{7 - 4x}{x-1} \]
The denominator is:
\[ 2 - \frac{2}{x-1} = \frac{2(x-1) - 2}{x-1} = \frac{2x - 2 - 2}{x-1} = \frac{2x - 4}{x-1} = \frac{2(x-2)}{x-1} \]
- Final Simplification
Now, we divide the simplified numerator by the simplified denominator:
\[ \frac{\frac{7-4x}{x-1}}{\frac{2(x-2)}{x-1}} \]
This is equivalent to multiplying the numerator by the reciprocal of the denominator:
\[ \frac{7-4x}{x-1} \cdot \frac{x-1}{2(x-2)} \]
Since \(x
eq 1\), we can cancel the common factor of \(x-1\):
\[ \frac{7-4x}{2(x-2)} = \frac{-4x+7}{2(x-2)} \]
Comparing this result to the given options, it corresponds to the second choice.
Examples
Simplifying complex expressions is crucial in fields like electrical engineering, where complex circuits involving resistors or capacitors often lead to complex fractions when calculating equivalent resistance or impedance. For instance, determining the total resistance in a network requires reducing these expressions to find a single, manageable value. Mastering this algebraic skill helps in solving for unknown physical properties in such systems efficiently.