Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated operations and simplify. \(\frac{2}{6 y-2}+\frac{y}{9 y^{2}-1}\)

Step-by-Step Solution

Answer

The expression simplifies to \(\frac{4y+1}{(3y-1)(3y+1)}\).

Step 1: Simplify Each Denominator

The first fraction is \(\frac{2}{6y-2}\). We can factor the denominator: \(6y-2 = 2(3y-1)\). The fraction becomes \(\frac{2}{2(3y-1)}\). Simplifying gives \(\frac{1}{3y-1}\).The second fraction is \(\frac{y}{9y^2-1}\). We can factor the denominator using the difference of squares: \(9y^2-1 = (3y-1)(3y+1)\). The fraction becomes \(\frac{y}{(3y-1)(3y+1)}\).

Step 2: Find a Common Denominator

Both fractions now have denominators involving \((3y-1)\). The least common denominator for the two fractions is \((3y-1)(3y+1)\).

Step 3: Rewrite Each Fraction with the Common Denominator

The first fraction \(\frac{1}{3y-1}\) is rewritten with the common denominator by multiplying numerator and denominator by \((3y+1)\), resulting in \(\frac{1 \cdot (3y+1)}{(3y-1)(3y+1)} = \frac{3y+1}{(3y-1)(3y+1)}\).The second fraction \(\frac{y}{(3y-1)(3y+1)}\) is already expressed with the common denominator.

Step 4: Combine the Fractions

Combine the numerators over the common denominator: \(\frac{3y+1}{(3y-1)(3y+1)} + \frac{y}{(3y-1)(3y+1)} = \frac{3y+1 + y}{(3y-1)(3y+1)}\).

Step 5: Simplify the Numerator

Add the terms in the numerator: \(3y + 1 + y = 4y + 1\). The expression becomes \(\frac{4y+1}{(3y-1)(3y+1)}\).

Step 6: Verify Simplification

Check if \(4y+1\) can be further simplified or factored with the denominator. Since there are no common factors, the expression \(\frac{4y+1}{(3y-1)(3y+1)}\) is the simplest form.

Key Concepts

Common Denominator

A common denominator is necessary when adding or subtracting fractions with different denominators. It allows us to combine the fractions into one single fraction. To find a common denominator, choose the smallest expression that both denominators can divide into evenly. This is known as the Least Common Denominator (LCD).
In our exercise, we dealt with \(\frac{1}{3y-1}\) and \(\frac{y}{(3y-1)(3y+1)}\). The LCD here is \( (3y-1)(3y+1) \).
Finding the common denominator involved:

• Recognizing that both fractions share \(3y-1\) as a factor.
• Using \( (3y-1)(3y+1) \) as it incorporates all factors present in both denominators.

This commonality made it easier to rewrite and combine the fractions. After adjusting each fraction to share the LCD, adding them together required only straightforward algebraic addition.

Factoring

Factoring simplifies algebraic expressions by breaking them down into simpler components, usually products of smaller expressions. In this context, knowing how to factor expressions helps simplify fractions.
For the first fraction, \(\frac{2}{6y-2}\), factor the denominator by finding common factors. Here, \(6y - 2\) can be factored as \(2(3y - 1)\), which greatly simplifies the fraction to \(\frac{1}{3y-1}\).
Factoring practiced here involved:

• Identifying common factors (as seen with \(6y-2\)).
• Applying special patterns like the difference of squares.

Recognizing these patterns can make handling algebraic expressions less complex, saving time and reducing errors in calculations.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations. Understanding and manipulating them is crucial in algebra. They allow complex problems to be broken down into manageable steps.
In this exercise, each fraction had algebraic expressions in their denominators that initially appeared difficult to handle.
The key steps to simplify included:

• Factoring these expressions (such as using the difference of squares).
• Finding a greater common denominator to ease combining them.

Handling algebraic expressions often requires identifying patterns, like factoring techniques, to reorganize them into easier pieces for further mathematical operations.

Difference of Squares

The difference of squares is a factoring technique used when two square terms are subtracted. The formula is simple: \(a^2 - b^2 = (a-b)(a+b)\). Recognizing this makes factoring easier when applicable.
In our problem, the denominator \(9y^2 - 1\) was factored using this method. By comparing to \(a^2 - b^2\), it turns into \( (3y-1)(3y+1) \).
This method involved:

• Identifying that both terms in \(9y^2 - 1\) are squares, with \(9y^2 = (3y)^2\) and \(-1 = -1^2\).
• Applying the difference of squares formula directly to simplify the expression.

Discovering and applying such techniques can drastically streamline working with algebra and make seemingly tough problems more manageable.