Problem 78 - Intermediate Algebra

Question

Perform the operations and simplify. $$ \frac{3 a^{3} b}{25 c d^{3}} \cdot \frac{5 c d^{2}}{6 a b} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{a^{2}}{10d} \).

Step 1: Set Up the Expression

Start by writing down the original expression to understand the terms involved. We have:\[\frac{3 a^{3} b}{25 c d^{3}} \cdot \frac{5 c d^{2}}{6 a b}\]This is a multiplication of two fractions.

Step 2: Multiply the Numerators

Multiply the numerators of both fractions together:\[3 a^{3} b \times 5 c d^{2} = 15 a^{3} b c d^{2}\]This will be the numerator of the new fraction.

Step 3: Multiply the Denominators

Multiply the denominators of both fractions together:\[25 c d^{3} \times 6 a b = 150 c a b d^{3}\]This will be the denominator of the new fraction.

Step 4: Combine and Simplify the Fraction

Combine the results from Steps 2 and 3 into a single fraction:\[\frac{15 a^{3} b c d^{2}}{150 c a b d^{3}}\]Now, simplify this fraction by canceling out common terms in the numerator and denominator. Start by simplifying the numerical coefficients:\[\frac{15}{150} = \frac{1}{10}\]Then, simplify the variable expressions:\[\frac{a^{3}}{a} = a^{2}, \quad \frac{b}{b} = 1, \quad \frac{c}{c} = 1, \quad \frac{d^{2}}{d^{3}} = \frac{1}{d}\]Thus, after canceling the common terms, the expression simplifies to:\[\frac{a^{2}}{10 d}\]

Key Concepts

Fractions Multiplication

Understanding fractions multiplication is key to solving many algebraic problems. In fractions, the multiplication process is straightforward and involves two main steps:

• Multiplying Numerators: Multiply the numerators of the fractions together to get the new numerator. This step combines the components above the fraction line. For example, in the given problem \[\frac{3 a^{3} b}{25 c d^{3}} \times \frac{5 c d^{2}}{6 a b}\]the numerators \(3 a^{3} b\) and \(5 c d^{2}\) are multiplied to get \(15 a^{3} b c d^{2}\).
• Multiplying Denominators: Similarly, multiply the denominators to form the new denominator. This is the part below the fraction line, so here \[25 c d^{3} \times 6 a b\]is done to get \(150 c a b d^{3}\).

This allows us to construct a new fraction that represents the product of the original two fractions. It's a critical skill necessary for moving forward in algebra.

Simplifying Algebraic Expressions

Simplifying algebraic expressions means making them easier to understand and solve, while maintaining their equivalency. The main focus here is reducing complexity by:

• Reducing Coefficients: For this exercise, you divide the numerical parts of a fraction (coefficients). So, divide both 15 (numerator) and 150 (denominator) by their greatest common factor, which results in \(\frac{1}{10}\).
• Simplifying Variables: We also simplify by comparing powers and canceling out one with the other. For instance, \(\frac{a^{3}}{a}\) simplifies to \(a^{2}\), reducing the ones with higher exponents. Similarly, cancel other variables: \(\frac{b}{b} = 1\), \(\frac{c}{c} = 1\), and \(\frac{d^{2}}{d^{3}} = \frac{1}{d}\).

This process turns a complex algebraic expression into a simpler, more manageable form like \(\frac{a^{2}}{10 d}\).

Canceling Common Terms

Canceling common terms is one of the most powerful simplification techniques in algebra. This involves identifying terms that appear both in the numerator and the denominator and removing them to simplify the expression. Here's how you do it:

• Identify Common Terms: Look for variables and numbers that are present both above and below the line. For example, in the expression \[\frac{15 a^{3} b c d^{2}}{150 c a b d^{3}}\]we see \(b\) appears in both, so it cancels out.
• Divide and Simplify: Take these common terms and apply division. Dividing \(b\) by \(b\) gives 1, effectively removing it from the expression. Similarly, applying this to other variables like \(c\) and \(d\) reduces the fraction greatly.
• Check Remaining Terms: After removing common factors, ensure all terms have been simplified and check the entire fraction again. The result should be a fraction with only the necessary, irreducible terms remaining, like \(\frac{a^{2}}{10 d}\).

Applying these steps correctly reduces complexity and makes solving easier.