Problem 12 - Algebra A Combined Function

Question

Perform each indicated operation. Simplify if possible. \(\frac{5}{y^{2}}-\frac{y}{2 y+1}\)

Step-by-Step Solution

Answer

The result is \(\frac{-y^3 + 10y + 5}{y^2(2y+1)}\).

Step 1: Identify the Problem

The task is to subtract two algebraic fractions: \(\frac{5}{y^{2}}\) and \(\frac{y}{2y+1}\). The key is to find a common denominator.

Step 2: Find a Common Denominator

The denominators are \(y^2\) and \(2y+1\). Since they have no common factors, multiply them to get the common denominator \(y^2(2y+1)\).

Step 3: Rewrite Each Fraction with the Common Denominator

To rewrite \(\frac{5}{y^2}\) with the common denominator \(y^2(2y+1)\), multiply the numerator and denominator by \((2y+1)\): \(rac{5(2y+1)}{y^2(2y+1)}\).Rewriting \(\frac{y}{2y+1}\) is done by multiplying the numerator and denominator by \(y^2\):\(\frac{y(y^2)}{y^2(2y+1)}\).

Step 4: Perform the Subtraction

Subtract the rewriten fractions: \(\frac{5(2y+1)}{y^2(2y+1)} - \frac{y^3}{y^2(2y+1)}\).Combine into one fraction:\(\frac{5(2y+1) - y^3}{y^2(2y+1)}\).

Step 5: Simplify the Expression

Expand and simplify the numerator: \[5(2y+1) = 10y + 5\]Thus the expression becomes:\(\frac{10y + 5 - y^3}{y^2(2y+1)}\) or \(\frac{-y^3 + 10y + 5}{y^2(2y+1)}\).

Step 6: Final Expression

The expression cannot be simplified further. The final answer is \(\frac{-y^3 + 10y + 5}{y^2(2y+1)}\).

Key Concepts

Common Denominator

Understanding the concept of a common denominator is crucial when dealing with algebraic fractions. When you have two fractions with different denominators, you need to find a way to bring them to a common ground before performing any operation such as addition or subtraction.
One effective strategy is to multiply each fraction by an appropriate expression that will make their denominators identical. In our exercise, the denominators were \(y^2\) and \((2y+1)\). Since these two denominators share no common factors, the simplest way to combine them is to multiply them together.
This gives us the common denominator \(y^2(2y+1)\). It's important to rewrite each fraction in terms of this common denominator, which involves multiplying both the numerator and the denominator by whatever factor is missing. This ensures that the value of the fraction is not changed, only its appearance is adjusted to a common form with the other fraction.

Subtraction of Fractions

Subtracting fractions is an operation that requires the fractions to have the same denominator. Once you've established a common denominator, you can easily subtract the numerators while keeping this common denominator constant.
In our example, after rewriting each fraction, the problem becomes subtracting \(\frac{5(2y+1)}{y^2(2y+1)} - \frac{y^3}{y^2(2y+1)}\). With the denominators matching, the subtraction focuses on the numerators: \(5(2y+1) - y^3\).
Remember, the subtraction of fractions depends only on the numerators when the denominators are the same, making it easier to manage and manipulate these expressions. Be careful to maintain the correct sign of each term, as it influences the final expression.

Simplifying Expressions

Simplifying expressions is often the final step in solving algebraic problems. After performing the subtraction, we end up with a new expression where the goal is to express it in its simplest form.
Let's consider the expression \(\frac{5(2y+1) - y^3}{y^2(2y+1)}\). First, you'll want to expand the terms in the numerator: \(5(2y+1)\) becomes \(10y + 5\). This changes our expression to \(\frac{10y + 5 - y^3}{y^2(2y+1)}\).
Simplifying does not always mean the expression will reduce drastically—it just means we ensure that it can't be reduced any further. Here, once the expansion and combination of like terms are completed, the expression \(\frac{-y^3 + 10y + 5}{y^2(2y+1)}\) is already in its simplest form, indicating that no further cancellation or factorization is possible.