Problem 10 - Algebra 2 and Trigonometry

Question

In \(3-20\) , perform the indicated additions or subtractions and write the result in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \frac{a+1}{3 a}+\frac{3}{2} $$

Step-by-Step Solution

Answer

The result is \(\frac{11a + 2}{6a}\), undefined for \(a = 0\).

Step 1: Find a Common Denominator

Identify the denominators in the given fractions. Here, the first fraction is \( \frac{a+1}{3a} \) and the second is \( \frac{3}{2} \). The denominators are \(3a\) and \(2\). The least common multiple of \(3a\) and \(2\) is \(6a\), because \(3a\cdot2=6a\).

Step 2: Convert Each Fraction to Have the Common Denominator

Convert \(\frac{a+1}{3a}\) to \( \frac{k (a+1)}{6a} \), where \(k\) is the factor needed to get to the common denominator. To do this, multiply both numerator and denominator by \(2\): \(\frac{2(a+1)}{6a}\). Convert \(\frac{3}{2}\) to \(\frac{3k}{6a}\), where \(k\) is also needed to reach \(6a\). Hence, multiply both numerator and denominator by \(3a\): \(\frac{9a}{6a}\).

Step 3: Add the Converted Fractions

Now that both fractions have the common denominator \(6a\), add the numerators together: \(\frac{2(a+1) + 9a}{6a}\). Simplify the numerator: \(2a + 2 + 9a = 11a + 2\).

Step 4: Simplify the Resulting Fraction

The final result of the addition is \(\frac{11a + 2}{6a}\). Check to see if there is a common factor in the numerator and denominator to simplify further. Since there is no common factor, the fraction is already in its simplest form.

Step 5: Identify Undefined Values for the Variables

Look at the denominators in the process. The original denominators are \(3a\) and \(2\), and the common denominator used is \(6a\). This is not defined when \(a=0\), because division by zero is undefined. Hence, the fraction is not defined for \(a = 0\).

Key Concepts

Common Denominator

To perform operations like addition or subtraction on fractions, they must share the same denominator, known as the common denominator. This process allows the fractions to be combined easily since the denominators match.

For example, if given fractions

• \(\frac{a+1}{3a}\)
• \(\frac{3}{2}\)

we first identify the denominators as \(3a\) and \(2\).
The smallest multiple they both share, also known as the least common denominator (LCD), is found by considering the least common multiple of the two quantities.

• \(3a\) and \(2\) have an LCM of \(6a\) because \(3a \times 2 = 6a\).

Using this common denominator, the fractions can be rewritten as having the same bottom number, which will facilitate addition or subtraction.
Understanding common denominators simplifies the process of dealing with multiple fractions.

Simplifying Fractions

After combining fractions with a common denominator, it's important to simplify them. This involves reducing the fraction to its smallest form possible by eliminating any common factors in the numerator and the denominator.

For example, adding fractions like

• \(\frac{2(a+1)}{6a}\)
• \(\frac{9a}{6a}\)

results in the fraction \(\frac{2(a+1) + 9a}{6a}\).
Expanding and combining the terms in the numerator gives us \(11a + 2\). The simplified fraction is then

• \(\frac{11a + 2}{6a}\).

To ensure it is in its simplest form, check for any common factors. In this case, there are none, so it cannot be simplified further.
Simplifying fractions makes them easier to understand and work with in further calculations.

Undefined Values

When dealing with fractions, it is crucial to recognize when a fraction may become undefined. A fraction becomes undefined if its denominator is zero, as division by zero is not possible.

In the given exercise, the original denominators are \(3a\) and \(2\). The common denominator used in the solution is \(6a\). Since this denominator includes the variable \(a\), we must identify where the denominator will be zero to understand undefined values.

• If \(6a = 0\), then \(a = 0\).

Therefore, the expression is undefined for \(a = 0\) as division by zero cannot occur.
Listing these undefined values is essential when solving problems involving fractions. It helps to avoid errors and ensures that expressions remain valid in their required domain.