Problem 41 - Algebra A Combined Function

Question

Perform each indicated operation. Simplify if possible. \(\frac{7}{(x+1)(x-1)}+\frac{8}{(x+1)^{2}}\)

Step-by-Step Solution

Answer

\(\frac{15x - 1}{(x+1)^2(x-1)}\)

Step 1: Identify a Common Denominator

First, look at the expressions in the denominators: \((x+1)(x-1)\) and \((x+1)^2\). The least common denominator (LCD) between both fractions is \((x+1)^2(x-1)\).

Step 2: Rewrite Each Fraction with the LCD

Rewrite \(\frac{7}{(x+1)(x-1)}\) and \(\frac{8}{(x+1)^2}\) with the common denominator \((x+1)^2(x-1)\).For \(\frac{7}{(x+1)(x-1)}\), multiply both the numerator and the denominator by \((x+1)\) to get \(\frac{7(x+1)}{(x+1)^2(x-1)}\).For \(\frac{8}{(x+1)^2}\), multiply both the numerator and the denominator by \((x-1)\) to get \(\frac{8(x-1)}{(x+1)^2(x-1)}\).

Step 3: Add the Fractions

Now that both fractions have the same denominator, add their numerators:\[ \frac{7(x+1)}{(x+1)^2(x-1)} + \frac{8(x-1)}{(x+1)^2(x-1)} = \frac{7(x+1) + 8(x-1)}{(x+1)^2(x-1)} \]

Step 4: Simplify the Numerator

Simplify the expression in the numerator:\[ 7(x+1) + 8(x-1) = 7x + 7 + 8x - 8 = 15x - 1 \]

Step 5: Write the Final Expression

Write the final simplified expression:\[ \frac{15x - 1}{(x+1)^2(x-1)} \]

Key Concepts

Least Common Denominator

When working with fractions, the Least Common Denominator (LCD) is crucial for performing operations like addition or subtraction. The least common denominator is the smallest expression that can be evenly divided by each of the original denominators in the fractions being considered. In this exercise, we are dealing with the fractions \( \frac{7}{(x+1)(x-1)} \) and \( \frac{8}{(x+1)^2} \).

To find their LCD, identify the highest power of each factor in the denominators. Here, we have factors \((x+1)\) and \((x-1)\). Specifically:

• The term \((x+1)\) is present as \((x+1)\) in the first denominator and as \((x+1)^2\) in the second denominator. Thus, \((x+1)^2\) is the LCD for this factor.
• For \((x-1)\), it's present in its simplest form in the first fraction, so \((x-1)\) becomes part of the LCD.

Thus, combining both, the LCD for these fractions is \((x+1)^2(x-1)\). By finding the LCD, you're setting the stage for making the denominators the same, which is necessary for adding the fractions. This ensures that the numerators can be combined straightforwardly.

Fraction Addition

Adding fractions, especially algebraic ones, involves some key steps, all relying heavily on having a common denominator. Once the LCD is identified, the next process is to write each fraction with this denominator.

In this case, the fractions are \( \frac{7}{(x+1)(x-1)} \) and \( \frac{8}{(x+1)^2} \) and we rewrite them:

• For \( \frac{7}{(x+1)(x-1)} \), multiply both the top and bottom by \((x+1)\), resulting in \( \frac{7(x+1)}{(x+1)^2(x-1)} \).
• For \( \frac{8}{(x+1)^2} \), multiply both the numerator and denominator by \((x-1)\), giving \( \frac{8(x-1)}{(x+1)^2(x-1)} \).

Now, with the denominators aligned as \((x+1)^2(x-1)\), you simply add the numerators:
\( 7(x+1) + 8(x-1) \). This results in a single fraction:\[ \frac{7(x+1) + 8(x-1)}{(x+1)^2(x-1)} \]

These steps ensure that the fractions are combined correctly, keeping operations smooth and manageable even in complex algebraic expressions.

Simplifying Expressions

Simplifying expressions is the final step after aligning fractions under a common denominator. Here, we're focused on the numerator since the denominator is already simplified. Simplifying the numerator makes the overall expression cleaner and more usable in further calculations or evaluations.

In the exercise at hand, you have \( 7(x+1) + 8(x-1) \) in the numerator. Breaking it down:

• Distribute \( 7(x+1) \) to get \( 7x + 7 \).
• Distribute \( 8(x-1) \) to obtain \( 8x - 8 \).
• Combine like terms: \( 7x + 8x \) results in \( 15x \), and \( 7 - 8 \) results in \(-1\).

Thus, the numerator simplifies to \( 15x - 1 \).

The final expression becomes:
\[ \frac{15x - 1}{(x+1)^2(x-1)} \]

Simplification is key in algebra as it yields the most reduced form of the expression, making it easier to interpret, manipulate or solve in subsequent mathematical operations.