- Factor the denominator of the first fraction: \(6n+4 = 2(3n+2)\).
- Rewrite and simplify the expression by canceling common factors: \(\frac{2n}{2(3n+2)} \cdot \frac{3n+2}{3n-2} = \frac{n}{3n+2} \cdot \frac{3n+2}{3n-2} = \frac{n}{3n-2}\).
- Identify the numerator of the simplified result \(\frac{n}{3n-2}\), which is \(n\).
- Final Answer: \(\boxed{n}\).
Explanation
- Problem Analysis
We are given the expression \(\left(\frac{2n}{6n+4}\right)\left(\frac{3n+2}{3n-2}\right)\). Our goal is to simplify this expression by identifying common factors in the numerators and denominators and then to determine the final numerator of the simplified fraction. The expression involves algebraic manipulation of rational expressions.
- Simplification and Result
To simplify the expression, we begin by factoring the denominator of the first fraction. Notice that \(6n+4\) shares a common factor of \(2\). We can rewrite this as:
\[6n+4 = 2(3n+2).\]
Now, substitute this back into the original expression:
\[\left(\frac{2n}{2(3n+2)}\right)\left(\frac{3n+2}{3n-2}\right).\]
In the first fraction, the numerator \(2n\) and the factor of \(2\) in the denominator can be simplified by cancelling the factor of \(2\):
\[\left(\frac{n}{3n+2}\right)\left(\frac{3n+2}{3n-2}\right).\]
Next, observe that the term \((3n+2)\) appears in both the denominator of the first fraction and the numerator of the second fraction. These terms can be cancelled out:
\[\frac{n}{\cancel{3n+2}} \cdot \frac{\cancel{3n+2}}{3n-2} = \frac{n}{3n-2}.\]
Thus, the simplified form of the given expression is \(\frac{n}{3n-2}\).
- Final Answer Selection
The simplified expression is \(\frac{n}{3n-2}\). The numerator of this fraction is \(n\). Therefore, the correct option is \(n\). The final answer is \(\boxed{n}\).
Examples
Simplifying rational expressions is a vital skill in algebra used whenever you need to streamline complex formulas. For instance, if you are designing a circuit where resistance depends on several variables, you might end up with a ratio of polynomials. By simplifying these expressions, you can more easily understand how a change in one parameter, like the number of components \(n\), affects the overall performance of the circuit. This makes your calculations more efficient and the model easier to analyze.