Problem 30 - Algebra 1: Concepts and Skills

Question

Write the difference in simplest form. $$ \frac{3}{6 b^{2}}-\frac{1}{4 b} $$

Step-by-Step Solution

Answer

The difference in simplest form is \(\frac{2 - b}{4b^{2}}\).

Step 1: Find the Common Denominator

The denominators here are \(6b^{2}\) and \(4b\). A common denominator would be a value that both these denominators can divide into. The least common denominator (LCD) is \(12b^{2}\). To achieve this, multiply the first fraction by \(2\) and the second fraction by \(3b\). This results in \(\frac{3*2}{6b^{2}*2} - \frac{1*3b}{4b*3b}\) = \(\frac{6}{12b^{2}} - \frac{3b}{12b^{2}}\).

Step 2: Subtract the Fractions

Now that the two fractions have the same denominator, subtract the numerators. This results in \(\frac{6 - 3b}{12b^{2}}\).

Step 3: Simplify the Resulting Fraction

The numerator and the denominator have common factor of 3. Divide the numerator and the denominator by this common factor to simplify the fraction. This yields \(\frac{2 - b}{4b^{2}}\).

Key Concepts

Least Common Denominator

In fraction arithmetic, the Least Common Denominator (LCD) is an essential concept when adding or subtracting fractions. The LCD is the smallest multiple that two or more denominators can divide into evenly. Finding the LCD allows us to rewrite fractions so that they have the same denominator, which in turn makes subtraction or addition possible.

To find the LCD of the fractions \(\frac{3}{6b^2}\) and \(\frac{1}{4b}\), we start by breaking down each denominator to their prime factors:

• \(6b^2 = 2 \times 3 \times b \times b\)
• \(4b = 2 \times 2 \times b\)

The LCD must include all prime factors from each denominator, the highest power of each, so:- \(2^2\) because it is the highest power of 2 occurring in the denominators.- \(3^1\) because it needs to appear for the first denominator.- \(b^2\) because it appears in its highest power in \(6b^2\).
The product is \(12b^2\), which is the LCD. By converting both fractions to a common denominator, we can easily subtract them.

Fraction Subtraction

Subtracting fractions is a straightforward task once you have a common denominator. It allows you to directly subtract the numerators while keeping the denominator constant. For the given fractions \(\frac{3}{6b^2}\) and \(\frac{1}{4b}\), after finding the least common denominator, we adjust the fractions:

• Multiply \(\frac{3}{6b^2}\) by \(\frac{2}{2}\) to get \(\frac{6}{12b^2}\)
• Multiply \(\frac{1}{4b}\) by \(\frac{3b}{3b}\) to get \(\frac{3b}{12b^2}\)

With both fractions now having the denominator \(12b^2\), you can subtract the numerators directly: \(6 - 3b\).
Therefore, the result of the subtraction is \(\frac{6 - 3b}{12b^2}\). This equation demonstrates how working with an LCD streamlines the subtraction of fractions.

Simplifying Fractions

Once you have performed the subtraction of fractions, the next step is to simplify the resulting fraction by dividing both the numerator and the denominator by their greatest common factor (GCF). Simplifying reduces fractions to their simplest form, making them easier to understand and work with.

From the subtraction, we have the fraction \(\frac{6 - 3b}{12b^2}\). Let's simplify it:

• The numerator \(6 - 3b\) can be factored by taking out a 3: \(3(2 - b)\)
• The denominator \(12b^2\) can be expressed as \(3 \times 4b^2\)

The GCF of both the numerator and the denominator is 3. Dividing each by 3 simplifies the fraction:\[ \frac{6 - 3b}{12b^2} = \frac{3(2 - b)}{3 \times 4b^2} = \frac{2-b}{4b^2} \]
This simplified form reveals the essence of the fraction, free from any unnecessary complexity. Simplified fractions are clearer and more manageable for further mathematical operations or analysis.