Problem 39 - Algebra for College Students

Question

Add or subtract the rational expressions as indicated. Be sure to express your answers in simplest form. $$ \frac{3}{n^{2}}-\frac{2}{5 n}+\frac{4}{3} $$

Step-by-Step Solution

Answer

\(\frac{20n^2 - 6n + 45}{15n^2}\)

Step 1: Identify the Denominators

In the expression \( \frac{3}{n^{2}} - \frac{2}{5n} + \frac{4}{3} \), we have three denominators: \(n^2\), \(5n\), and \(3\). To add or subtract these fractions, we need to find a common denominator.

Step 2: Find the Least Common Denominator (LCD)

The least common denominator of \(n^2\), \(5n\), and \(3\) is the least common multiple of these terms. The LCD is \(15n^2\).

Step 3: Rewrite Each Fraction with the LCD

Rewrite each fraction with the common denominator, \(15n^2\). This gives: \[ \frac{3}{n^2} = \frac{3 imes 15}{n^2 imes 15} = \frac{45}{15n^2} \]\[ \frac{2}{5n} = \frac{2 imes 3n}{5n imes 3n} = \frac{6n}{15n^2} \]\[ \frac{4}{3} = \frac{4 imes 5n^2}{3 imes 5n^2} = \frac{20n^2}{15n^2} \]

Step 4: Combine the Fractions

Now we add and subtract the fractions with the common denominator:\[ \frac{45}{15n^2} - \frac{6n}{15n^2} + \frac{20n^2}{15n^2} = \frac{45 - 6n + 20n^2}{15n^2} \]

Step 5: Simplify the Expression

The numerator \(45 - 6n + 20n^2\) is already simplified as the terms cannot be further combined. Thus, the expression in simplest form is:\[ \frac{20n^2 - 6n + 45}{15n^2} \]

Key Concepts

Least Common Denominator

When working with fractions, especially rational expressions, finding a common basis for comparison is crucial. This is where the concept of the Least Common Denominator (LCD) comes in. The least common denominator is the smallest multiple that is common to all denominators involved. This allows us to perform operations like addition or subtraction on fractions by converting them to equivalent fractions with the same denominator.
In the given problem, the denominators are \(n^2\), \(5n\), and \(3\). To find the LCD, we need to take each distinct factor with the highest power across these denominators:

• The factor \(n\) appears as \(n^2\) and \(5n\), so \(n^2\) will be included.
• The factor \(5\) in \(5n\) is included.
• The constant \(3\) is included from the last term.

Thus, the LCD becomes \(15n^2\), which is a combination of these factors. This unified denominator is then used to rewrite each fraction so they can be easily added or subtracted. Identifying the LCD correctly is key to successfully managing rational expressions.

Simplifying Expressions

Once you establish a common denominator, the next step in dealing with rational expressions is to simplify them. Simplification involves converting each term to a common foundation and then reducing any combined terms to their most basic form, if possible.
In our example, the fractions are rewritten as:

• \(\frac{3}{n^2} = \frac{45}{15n^2}\)
• \(\frac{2}{5n} = \frac{6n}{15n^2}\)
• \(\frac{4}{3} = \frac{20n^2}{15n^2}\)

The numerators were adjusted to reflect the change in each fraction over the common denominator, \(15n^2\), ensuring that each fraction is equivalent to its original form. Now, the fractions are ready to be added or subtracted. The step before calculation ensures that the fractions are expressed in a way that simplifies the algebraic manipulation to follow. Simplifying is crucial for making expressions manageable and for preventing mistakes during the calculation process.

Adding and Subtracting Fractions

With the fractions sharing a common denominator, the actual process of adding or subtracting them becomes far simpler. This is because you only need to focus on the numerators while keeping the denominator constant.
For the task at hand, you have:

• \(\frac{45}{15n^2}\)
• \(- \frac{6n}{15n^2}\)
• \(+ \frac{20n^2}{15n^2}\)

Now, you can perform the operation straightforwardly:\[\frac{45}{15n^2} - \frac{6n}{15n^2} + \frac{20n^2}{15n^2} = \frac{45 - 6n + 20n^2}{15n^2}\]The result is still shown over the common denominator of \(15n^2\). Further simplification, in this case, cannot combine terms any further as they represent different orders of \(n\).
This step consolidates the terms into a single fraction, giving you a final answer that is not only correct but also expressed in its simplest form. Understanding this step is fundamental as it ties together the usefulness of finding an LCD and rewriting terms for coherent arithmetic operations among fractions.