What is the product?
$\frac{4 n}{4 n-4} \cdot \frac{n-1}{n+1}$
To simplify the expression \(\frac{4n}{4n-4} \cdot \frac{n-1}{n+1}\), we first analyze the individual components. The expression is a product of two rational functions. For the expression to be well-defined, we assume \(n eq 1\) (to avoid division by zero in the first fraction) and \(n eq -1\) (to avoid division by zero in the second fraction). Our goal is to simplify this product by factoring and canceling common terms.
The denominator of the first fraction is \(4n-4\). We can factor out the common constant \(4\):
\[4n-4 = 4(n-1)\]
Substituting this back into the first fraction, we get:
\[\frac{4n}{4(n-1)}\]
We can simplify this by canceling the factor of \(4\) in both the numerator and the denominator:
\[\frac{4n}{4(n-1)} = \frac{n}{n-1}\]
Now, we multiply this simplified result by the second fraction:
\[\frac{n}{n-1} \cdot \frac{n-1}{n+1}\]
By combining the fractions, we get:
\[\frac{n(n-1)}{(n-1)(n+1)}\]
Since \(n
eq 1\), the factor \((n-1)\) is not zero, so we can cancel it from both the numerator and the denominator:
\[\frac{n \cancel{(n-1)}}{\cancel{(n-1)}(n+1)} = \frac{n}{n+1}\]
Comparing our result, \(\frac{n}{n+1}\), with the provided options, we see that it matches the second option. Thus, the simplified form of the given expression is \(\frac{n}{n+1}\). Therefore, the final answer is \(\boxed{\frac{n}{n+1}}\).
Simplifying rational expressions is essential in fields like physics and engineering, where complex formulas often need to be reduced to their simplest forms to perform calculations easily. For example, if you are modeling the speed of an object subject to two different resistive forces represented by fractions, you might find that the total effective resistance simplifies significantly when you multiply and reduce the terms, just as we did here. This allows for quicker analysis and a better understanding of how the variable, such as time or mass, influences the overall system.