Problem 57 - Introductory Algebra for College Students

Question

Divide as indicated. $$\frac{2 y^{2}-128}{y^{2}+16 y+64}+\frac{y^{2}-6 y-16}{3 y^{2}+30 y+48}$$

Step-by-Step Solution

Answer

The simplified form of the given expression is \(\frac{7y - 56}{3(y + 8)}\).

Step 1: Factorization

We begin by factorizing the numerator and denominator of the given fractions. This leads to: \(\frac{(2y - 16)(y + 8)}{(y + 8)^2} + \frac{(y - 8)(y + 2)}{3(y + 8)(y + 2)}.\) You'll notice that there is a common term in both fractions' denominators.

Step 2: Simplify the fractions

We simplify the fractions: \(\frac{2y - 16}{(y + 8)} + \frac{y - 8}{3(y + 8)}\). Here (y + 8) is the common denominator for both fractions.

Step 3: Combine the fractions

With the common denominator in place, it's possible to combine the two fractions into one. This results in \(\frac{2y - 16 + \frac{y - 8}{3}}{y + 8}\) Then to further simplify this fraction, clear the denominator in the fraction in the numerator by multiplying all terms by 3: \(\frac{6y - 48 + y - 8}{3(y + 8)}\)

Step 4: Simplify the result

Add the like terms in the numerator to get the final result: \(\frac{7y - 56}{3(y + 8)}\)

Key Concepts

Factorization

Factorization is the process of breaking down an expression into a product of its simpler factors. It is a crucial skill in algebra as it simplifies expressions, making them easier to solve or combine. For the expression in our exercise, we begin by examining each term both in numerators and denominators.

• The first fraction's numerator, \(2y^2 - 128\), can be simplified by factoring out common terms. Notice that 2 is a common factor: \(2(y^2 - 64)\).
• Further, \(y^2 - 64\) is a difference of squares, which factors as \((y - 8)(y + 8)\).

Thus, the complete factorization for the numerator becomes \((2y - 16)(y + 8)\).

• For the denominator, \(y^2 + 16y + 64\), we must find two numbers that multiply to 64 and add up to 16. These numbers are 8 and 8, giving us \((y + 8)^2\).

By understanding factorization, you transform more complex expressions into manageable products that hint at potential simplifications.

Simplifying Fractions

Simplifying fractions involves reducing the fraction to its simplest form, where the numerator and denominator share no common factors other than 1. This process is key when dealing with algebraic fractions because it makes calculations more straightforward.

Once we have the factorized version of the fractions from the exercise, simplification is the natural next step.

• In the original expression, \(\frac{(2y - 16)(y + 8)}{(y + 8)^2}\), the factor \((y + 8)\) appears in both the numerator and the denominator.
• Thus, you can divide both by \((y + 8)\), simplifying the expression directly to \(\frac{2y - 16}{y + 8}\).

Always check for common factors in both parts of the fraction to ensure it's in its simplest form. Simplifying before proceeding with the solution helps avoid dealing with unnecessarily cumbersome expressions.

Common Denominators

To add or subtract fractions, their denominators must be the same. The common denominator creates a shared baseline, allowing the numerators to be combined easily. This method is fundamental in algebra, as our exercise demonstrates naturally.

In the exercise, once factorization and simplification identified the shared factor of \((y + 8)\) in both fractions' denominators:

• Our goal was to consolidate terms under a single shared denominator, which simplifies combining the fractions.
• After simplification, both fractions shared the common denominator \((y + 8)\), leading us to easily combine them: \(\frac{2y - 16 + \frac{y - 8}{3}}{y + 8}\).

Using the common denominator makes further simplification possible by focusing on the numerators. This foundational concept ensures algebraic fractions become simpler equations that are more straightforward to handle and solve.