Problem 60 - Algebra A Combined Function

Question

Perform each indicated operation. Simplify if possible. \(\frac{6}{5 y^{2}-25 y+30}-\frac{2}{4 y^{2}-8 y}\)

Step-by-Step Solution

Answer

\(\frac{-10y^2 + 74y - 60}{20y(y - 2)(y - 3)}\)

Step 1: Factor each denominator

To make the denominators ready for adding or subtracting fractions, we factor them. Begin with the first fraction: the denominator is \(5y^2 - 25y + 30\). This can be factored into \((5y - 10)(y - 3)\). Now for the second fraction: the denominator is \(4y^2 - 8y\). This can be factored as \(4y(y - 2)\).

Step 2: Find a common denominator

The common denominator is the least common multiple of \((5y - 10)(y - 3)\) and \(4y(y - 2)\). To find this, multiply the distinct factors together: \(20y(y - 2)(y - 3)\).

Step 3: Rewrite each fraction with the common denominator

Rewrite \(\frac{6}{(5y - 10)(y - 3)}\) as \(\frac{6(4y)}{20y(y - 2)(y - 3)}\) and \(\frac{2}{4y(y - 2)}\) as \(\frac{2(5y - 10)(y - 3)}{20y(y - 2)(y - 3)}\).

Step 4: Subtract the numerators

Combine the fractions: \(\frac{6(4y)}{20y(y - 2)(y - 3)} - \frac{2(5y - 10)(y - 3)}{20y(y - 2)(y - 3)}\) and subtract the numerators: \(6(4y) - 2(5y - 10)(y - 3)\).

Step 5: Simplify the expression

Calculate the new numerator by distributing: \(24y - [10y^2 - 50y + 60]\), which simplifies to \(-10y^2 + 74y - 60\). The overall expression is \(\frac{-10y^2 + 74y - 60}{20y(y - 2)(y - 3)}\).

Step 6: Check for further simplification

Check if the new numerator \(-10y^2 + 74y - 60\) can be factored for further simplification. In this case, it does not factor nicely into terms that will cancel with the denominator, leaving the expression simplified as it is.

Key Concepts

Factoring Polynomials

To work with algebraic fractions, it's crucial to understand polynomial factoring. Factoring polynomials involves writing the expression as a product of its simpler factors. This is a foundational step when dealing with fractions, as it prepares the expression for operations like adding or subtracting.

In our exercise, we started by factoring the denominator of each fraction. For the first polynomial, we had to factor \(5y^2 - 25y + 30\). This expression was broken down into two binomials: \((5y - 10)(y - 3)\). Similarly, the second polynomial, \(4y^2 - 8y\), was factored as \(4y(y - 2)\).

Without factoring, attempting to find a common denominator or simplifying would be incredibly complex if not outright impossible. Factoring strips the polynomial down to its core components, making further steps manageable.

Common Denominator

Finding a common denominator is essential for adding or subtracting fractions, whether numerical or algebraic. The common denominator lets us combine the fractions into a single expression because the denominators must match.

For algebraic fractions, we look for the least common multiple (LCM) of the denominators. In our problem, after factoring, we identified the LCM as \((5y - 10)(y - 3)\) and \(4y(y - 2)\). By multiplying these distinct factors, we found the common denominator to be \(20y(y - 2)(y - 3)\).

This step ensures both fractions can be rewritten with the same denominator, making subtraction possible. It involves creatively working with both numerical and algebraic factors to find the smallest shared denominator.

Subtracting Fractions

Once the denominators are identical, you can subtract fractions by directly subtracting the numerators. This method simplifies complex expressions to manageable forms.

In our example, after rewriting each fraction with the common denominator \(20y(y - 2)(y - 3)\), we had the expressions \(\frac{6(4y)}{20y(y - 2)(y - 3)}\) and \(\frac{2(5y - 10)(y - 3)}{20y(y - 2)(y - 3)}\).

To subtract these fractions, we computed \(6(4y) - 2(5y - 10)(y - 3)\). By distributing and combining like terms, the result was a new expression: \(-10y^2 + 74y - 60\).

Finally, we recognized that the numerator could not be further factored in a way that cancels with the denominator. Thus, the fraction remained as \(\frac{-10y^2 + 74y - 60}{20y(y - 2)(y - 3)}\). Breaking it down into simple steps helps reveal the beauty of algebraic manipulation and the logical process behind solving such problems.