Problem 8 - Basic Technical Mathematics with Calculus

Question

Multiply the numerator and the denominator of each fraction by the given factor and obtain an equivalent fraction. $$\frac{2 x^{2} y}{3 n} \quad\left(\text { by } 2 x n^{2}\right)$$

Step-by-Step Solution

Answer

The equivalent fraction is \( \frac{4x^3yn^2}{6xn^3} \).

Step 1: Identify the Original Fraction

The original fraction given is \( \frac{2x^2 y}{3n} \). This fraction has the numerator \( 2x^2 y \) and the denominator \( 3n \).

Step 2: Determine the Multiplying Factor

The fraction needs to be multiplied by \( 2xn^2 \) to find an equivalent fraction. This means both the numerator and the denominator need to be multiplied by \( 2xn^2 \).

Step 3: Multiply the Numerator

Multiply the numerator \( 2x^2 y \) by \( 2xn^2 \). This gives:\[(2x^2y) \times (2xn^2) = 4x^3yn^2\]

Step 4: Multiply the Denominator

Multiply the denominator \( 3n \) by \( 2xn^2 \). This gives:\[(3n) \times (2xn^2) = 6xn^3\]

Step 5: Form the Equivalent Fraction

Combine the new numerator and denominator to form the equivalent fraction:\[\frac{4x^3yn^2}{6xn^3}\]

Key Concepts

Equivalent Fractions

In mathematics, equivalent fractions are different fractions that represent the same value or proportion. Imagine slicing a delicious pie. Whether you have two quarters or one half, you still enjoy the same amount of pie! That's how equivalent fractions work. To create an equivalent fraction, we multiply or divide both the numerator and the denominator of a fraction by the same non-zero number.

In the exercise provided, we have the fraction \( \frac{2x^2 y}{3n} \), and we multiply both the numerator and the denominator by the same factor \( 2xn^2 \). As a result, we form an equivalent fraction:

• New Numerator: \( 2x^2 y \times 2xn^2 = 4x^3yn^2 \)
• New Denominator: \( 3n \times 2xn^2 = 6xn^3 \)

This ensures the original value of the fraction remains unchanged, leading us to our new equivalent fraction \( \frac{4x^3yn^2}{6xn^3} \).

Understanding equivalent fractions helps to simplify calculations and compare different fractions easily.

Algebraic Fractions

Algebraic fractions are just like regular fractions, but they contain algebraic expressions in their numerators, denominators, or both. These can include variables like \( x \) or \( y \) that make them a bit more intriguing. For example, \( \frac{2x^2y}{3n} \) is an algebraic fraction because both the numerator and the denominator include such variables and terms.

When working with algebraic fractions, similar techniques used for numerical fractions apply. However, it involves a clear understanding of how to handle variables and perform operations like multiplication or division with expressions. In our exercise:

• Numerator initially: \( 2x^2y \)
• Denominator initially: \( 3n \)

To multiply and find the equivalent fraction, we carefully distribute and apply the factor \( 2xn^2 \) to both parts. This requires multiplying each term properly to maintain the integrity of the algebraic expression. Thus, ensuring you handle these operations correctly is essential for simplifying and solving algebraic fractions.

Algebraic fractions are handy tools in mathematics, often used in equations and expressions across various problem-solving scenarios.

Polynomial Multiplication

Polynomial multiplication involves expanding and simplifying expressions that can result in new polynomial expressions. Think of it like expanding an elastic rubber band; you're stretching it out to see all its parts! Polynomials can contain one or more terms, like our example numerator \( 2x^2y \). When multiplying by another polynomial, each term must be multiplied appropriately.

In our practice exercise, we multiplied \( 2x^2y \) by \( 2xn^2 \), expanding into:

• \( 2x^2y \times 2 = 4x^2y \)
• \( 4x^2y \times xn^2 = 4x^3yn^2 \)

The product \( 4x^3yn^2 \) is what you get when each term of one polynomial is multiplied by every term in the other.

This multiplication is particularly crucial for handling expressions in algebra, making it easier to expand, solve, or simplify equations. Understanding how to multiply polynomials accurately enables you to tackle more complex mathematical problems, revealing the power behind these expressions.