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Chapter 7: Problem 40

Divide, and then simplify, if possible. \(\frac{3}{a} \div \frac{9}{5}\)

Short Answer

Expert verified
\( \frac{3}{a} \div \frac{9}{5} = \frac{5}{3a} \).

Step by step solution

01

Rewrite the Division as Multiplication

To simplify the expression \( \frac{3}{a} \div \frac{9}{5} \), rewrite the division as multiplication by the reciprocal of the divisor. This means we will multiply \( \frac{3}{a} \) by the reciprocal of \( \frac{9}{5} \). The reciprocal of \( \frac{9}{5} \) is \( \frac{5}{9} \). Therefore, our expression becomes \( \frac{3}{a} \times \frac{5}{9} \).
02

Perform the Multiplication

Now, multiply the numerators and denominators separately in the expression \( \frac{3}{a} \times \frac{5}{9} \). The numerator will be \( 3 \times 5 = 15 \) and the denominator will be \( a \times 9 = 9a \). So, the result of multiplication is \( \frac{15}{9a} \).
03

Simplify the Fraction

To simplify the fraction \( \frac{15}{9a} \), find the greatest common divisor (GCD) of 15 and 9. The GCD is 3. Divide both the numerator and the denominator by 3. This yields \( \frac{15 \div 3}{9 \div 3a} = \frac{5}{3a} \). Thus, the simplified form of the expression is \( \frac{5}{3a} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Division of Fractions
Dividing fractions involves a unique twist compared to whole numbers or decimals. Instead of simply dividing as usual, in fractions, the process is based on multiplying with the reciprocal. Here鈥檚 the step-by-step guide to make division of fractions as simple as possible:
  • Step 1: Start by identifying the fraction that you need to divide. This is called the dividend.
  • Step 2: Recognize the fraction you're dividing by, known as the divisor.
  • Step 3: Find the reciprocal of the divisor. To find a reciprocal, essentially flip the fraction upside down, swapping its numerator and denominator.
  • Step 4: Change the division problem into a multiplication problem by multiplying with the reciprocal.
By turning the division into multiplication, you eliminate the complexity of division, using a more straightforward operation instead. As described in the exercise, when handling \( \frac{3}{a} \div \frac{9}{5} \), the division turns into \( \frac{3}{a} \times \frac{5}{9} \), making it more manageable to solve.
Simplifying Fractions
Simplifying fractions is a powerful tool to make problems easier and solutions cleaner. It鈥檚 about reducing the fraction to its simplest form, without changing its value. Here鈥檚 how to simplify fractions effectively:
  • Identify the GCD: To simplify, first find the greatest common divisor (GCD) of the numerator and the denominator. The GCD is the largest number that can exactly divide both numbers.
  • Divide: Use the GCD to divide both the numerator and the denominator. This reduces the fraction to its simplest form.
  • Check Your Work: After dividing, it's important to check whether further simplification is possible by ensuring no common factors remain.
In the exercise, the fraction \( \frac{15}{9a} \) is simplified by recognizing that both 15 and 9 are divisible by 3, yielding the fraction \( \frac{5}{3a} \). This process not only makes the solution more elegant but also more straightforward to interpret.
Reciprocal in Mathematics
The concept of a reciprocal is essential in dividing fractions and has a broad range of applications in mathematics. Understanding reciprocals can often make complex operations much simpler.
  • Finding a Reciprocal: To find the reciprocal of a fraction, invert it. For example, the reciprocal of \( \frac{9}{5} \) is \( \frac{5}{9} \).
  • Properties of Reciprocals: When you multiply a number by its reciprocal, the result is always 1, because the numerator and the denominator cancel each other out.
  • Application in Division: Reciprocals transform division into multiplication, sidestepping the division operation, which can be more complicated with fractions.
By transforming \( \frac{3}{a} \div \frac{9}{5} \) into \( \frac{3}{a} \times \frac{5}{9} \), the problem simplifies, demonstrating how reciprocals streamline calculations in algebraic expressions.

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