Problem 34 - Elementary and Intermediate Algebra

Question

Perform the operations. Simplify, if possible. $$ \frac{2 y}{5 y-1}-\frac{2 y}{3 y+2} $$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{-2y(2y-3)}{(5y-1)(3y+2)}\).

Step 1: Establish a Common Denominator

To perform the subtraction, we need a common denominator. The denominators are \(5y - 1\) and \(3y + 2\). The common denominator is their product: \((5y - 1)(3y + 2)\).

Step 2: Rewrite Each Fraction

Express each fraction with the common denominator.\[\frac{2y}{5y-1} = \frac{2y(3y+2)}{(5y-1)(3y+2)}\]\[\frac{2y}{3y+2} = \frac{2y(5y-1)}{(5y-1)(3y+2)}\]

Step 3: Perform Subtraction

Subtract the rewritten fractions:\[\frac{2y(3y+2) - 2y(5y-1)}{(5y-1)(3y+2)}\] Simplifying the numerator:\(2y(3y+2) = 6y^2 + 4y\) and \(2y(5y-1) = 10y^2 - 2y\). Subtract these:\[6y^2 + 4y - (10y^2 - 2y) = 6y^2 + 4y - 10y^2 + 2y = -4y^2 + 6y\]

Step 4: Simplify the Result

Combine the terms in the numerator:\[\frac{-4y^2 + 6y}{(5y-1)(3y+2)}\]Factor the numerator:\[ -2y(2y - 3) \] for simplification purposes.Thus, the expression becomes:\[\frac{-2y(2y - 3)}{(5y-1)(3y+2)}\]

Key Concepts

Common Denominator

When dealing with fraction subtraction, one essential first step is to establish a common denominator. This part can often feel like a puzzle, yet once understood, makes working with fractions much smoother.
Each fraction has a denominator, the number located below the fraction bar, which tells us how many parts the whole is divided into. When these denominators are different, as they are in our example with \(5y - 1\) and \(3y + 2\), we need to find a common denominator to "communicate" clearly between the fractions by making sure they refer to the same size whole.
A common denominator is simply the least common multiple of the denominators of the fractions you're working with. In our scenario, multiplying the two existing denominators, giving us \( (5y - 1)(3y + 2) \), serves this purpose beautifully.
This allows for each fraction to be expressed as parts of this same, equivalent whole, paving the way for the next steps in your arithmetic operation.

Fraction Subtraction

Subtraction of fractions involves rewriting each fraction with the common denominator, which allows you to directly subtract the numerators. This is crucial because, similar to breaking chocolate bars into the same number of pieces, only then can you logically and consistently take pieces away. Here's how the subtraction process unfolds:
1. **Express Each Fraction**: Start by adjusting each fraction so they both share the common denominator you previously identified ((5y - 1)(3y + 2)). For instance, \(\frac{2y}{5y-1}\) becomes \(\frac{2y(3y+2)}{(5y-1)(3y+2)}\) and \(\frac{2y}{3y+2}\) converts into \(\frac{2y(5y-1)}{(5y-1)(3y+2)}\).
2. **Perform Subtraction**: Now subtract the numerators: \(2y(3y+2) \) minus \(2y(5y-1)\). After clearing brackets and reorganizing terms, the subtraction becomes manageable, leaving you with a single expression \(\frac{2y(3y+2) - 2y(5y-1)}{(5y-1)(3y+2)}\).
This brings you to one of the heartbeats of fraction arithmetic: practicing clear systematic manipulation of numerators and denominators, patiently setting the stage for a clean simplification.

Simplifying Expressions

The core goal of simplifying expressions is to tidy up what often starts as complex to a neat, recognizable form. Together with common denominators and fraction subtraction, it plays a vital role in clearly unfolding algebraic ideas. Let’s simplify some expressions!
1. **Simplify the Numerator**: Begin with focusing on the numerator. You have done the arithmetic to find \(-4y^2 + 6y\), which came from subtracting \(10y^2 - 2y\) from \(6y^2 + 4y\).
2. **Factor if Possible**: Look to factor the resulting numerator if it makes it simpler. Here, you can pull out a common factor, which is \(-2y\), giving you \(-2y(2y - 3)\). This not only organizes the expression but also reveals its core terms.
3. **Conclusion**: The expression neatly boils down to \(\frac{-2y(2y - 3)}{(5y-1)(3y+2)}\). You've taken what initially might have appeared complex and returned a more streamlined form which is easy to understand and further work with.
Simplifying is like resolving a puzzle, and when each piece falls into place, it illuminates the elegance of algebra more brightly.