Verified Answer :

  • Combine the terms in the denominator by finding a common denominator \(yx^2\), resulting in \(\frac{x^2+5y}{yx^2}\).
  • Rewrite the complex fraction as division, \(\frac{3}{y^2x} \div \frac{x^2+5y}{yx^2}\).
  • Convert the division to multiplication by the reciprocal, \(\frac{3}{y^2x} \cdot \frac{yx^2}{x^2+5y}\).
  • Simplify by canceling the common terms \(x\) and \(y\) to reach the final form \(\boxed{\frac{3x}{yx^2 + 5y^2}}\).

Explanation

  1. Identifying the Components of the Complex Fraction

To simplify the complex fraction \(\frac{\frac{3}{y^2 x}}{\frac{1}{y} + \frac{5}{x^2}}\), we first identify its components. The numerator is \(\frac{3}{y^2 x}\) and the denominator is \(\frac{1}{y} + \frac{5}{x^2}\). We assume that the variables \(x\) and \(y\) are non-zero, ensuring the expressions are well-defined.

  1. Simplifying the Denominator

Next, we focus on simplifying the denominator \(\frac{1}{y} + \frac{5}{x^2}\). To add these two terms, we need a common denominator, which is \(yx^2\). We rewrite each fraction:
\(\frac{1}{y} = \frac{x^2}{yx^2}\) and \(\frac{5}{x^2} = \frac{5y}{yx^2}\).
Summing these gives:
\(\frac{x^2 + 5y}{yx^2}\).

  1. Converting Division to Multiplication

Now we can rewrite the entire expression as a division of two fractions:
\[ \frac{3}{y^2 x} \div \frac{x^2 + 5y}{yx^2} \]
To divide by a fraction, we multiply by its reciprocal:
\[ \frac{3}{y^2 x} \times \frac{yx^2}{x^2 + 5y} \]

  1. Simplifying the Final Expression

Finally, we multiply the numerators and denominators:
\[ \frac{3 \times y \times x^2}{y^2 \times x \times (x^2 + 5y)} \]
Next, we simplify by canceling common factors. Notice that \(y\) in the numerator cancels with \(y^2\) in the denominator (leaving \(y\) in the denominator) and \(x\) in the denominator cancels with \(x^2\) in the numerator (leaving \(x\) in the numerator).
\[ \frac{3x}{y(x^2 + 5y)} = \boxed{\frac{3x}{yx^2 + 5y^2}} \]

Examples

In physics or engineering, you might encounter complex formulas describing how components interact. For example, if you are calculating the equivalent resistance or focal length in a circuit or lens system where variables are embedded within complex fractions, simplifying these expressions allows you to better understand the relationship between your parameters. Reducing a complex fraction like this one simplifies the computational workload for further analysis.