- Identify the common denominator of the inner fractions, which is \(y\).
- Multiply both the numerator and denominator by \(y\) to eliminate the inner fractions.
- Distribute \(y\) to simplify both parts: \(\frac{(2 \cdot y) - (\frac{1}{y} \cdot y)}{(3 \cdot y) + (\frac{1}{y} \cdot y)} = \frac{2y-1}{3y+1}\).
- The equivalent expression is \(\boxed{\frac{2y-1}{3y+1}}\).
Explanation
- Problem Analysis and Strategy
To simplify the complex fraction \(\frac{2-\frac{1}{y}}{3+\frac{1}{y}}\), we observe the inner fractions \(\frac{1}{y}\). Since \(y\) is a common denominator for these inner fractions, we can simplify the expression by multiplying both the numerator and the denominator of the main fraction by \(y\). We must assume \(y
eq 0\) for the expression to be defined, and \(3+\frac{1}{y}
eq 0\), which implies \(y
eq -\frac{1}{3}\).
- Step-by-Step Simplification
We multiply the entire fraction by \(\frac{y}{y}\), which is equivalent to 1:
\[\frac{2-\frac{1}{y}}{3+\frac{1}{y}} \cdot \frac{y}{y} = \frac{(2-\frac{1}{y}) \cdot y}{(3+\frac{1}{y}) \cdot y}\]
Now, distribute the \(y\) to each term in the numerator and the denominator:
Numerator: \(2 \cdot y - \left(\frac{1}{y}\right) \cdot y = 2y - 1\)
Denominator: \(3 \cdot y + \left(\frac{1}{y}\right) \cdot y = 3y + 1\)
Thus, the simplified form is \(\frac{2y-1}{3y+1}\).
- Comparison and Final Result
Comparing our result \(\frac{2y-1}{3y+1}\) with the given choices:
- \(\frac{3y+1}{2y-1}\)
- \(\frac{(2y-1)(3y+1)}{y^2}\)
- \(\frac{y^2}{(2y-1)(3y+1)}\)
- \(\frac{2y-1}{3y+1}\)
The simplified expression matches the fourth option. Therefore, the equivalent expression is \(\frac{2y-1}{3y+1}\).
Examples
Simplifying complex fractions is a fundamental skill in algebra that appears frequently when solving equations involving variables in denominators. For example, in electrical engineering, when analyzing circuits with multiple parallel resistors, you often encounter complex fractions where the total resistance is given by the reciprocal of the sum of reciprocals. Simplifying these expressions allows you to find a single, manageable formula to calculate current or voltage more easily. Mastery of this technique ensures you can quickly reduce complex equations to simpler, more intuitive forms.