Problem 25 - Prealgebra and Introductory Algebra

Question

Which expressions are equivalent to \(\frac{3}{y-5}-\frac{y-2}{y-5} ?\) (I) \(\frac{5-y}{y-5}\) (II) \(\frac{1-y}{y-5}\) (III) \(\frac{5-y}{2 y-10}\) (IV) \(-1\) (V) \(\frac{1-y}{-10}\)

Step-by-Step Solution

Answer

The expressions equivalent to \(\frac{3}{y-5}-\frac{y-2}{y-5}\) are (I) \(\frac{5-y}{y-5}\) and (III) \(\frac{5-y}{2 y-10}\).

Step 1: Simplify the given expression

Start with the expression \(\frac{3}{y-5}-\frac{y-2}{y-5}\). As both fractions have the same denominator (\(y - 5\)), they can be combined by subtracting the numerators. This results in: \(\frac{3 - (y - 2)}{y - 5}\), which simplifies to: \(\frac{5 - y}{y - 5}\).

Step 2: Compare with the options

The simplified expression \(\frac{5 - y}{y - 5}\) should now be compared with the given options: (I) \(\frac{5-y}{y-5}\), (II) \(\frac{1-y}{y-5}\), (III) \(\frac{5-y}{2 y-10}\), (IV) \(-1\), and (V) \(\frac{1-y}{-10}\). Comparing it with (I), it's clear that (I) \(\frac{5-y}{y-5}\) is equivalent to the given expression. However, (II) \(\frac{1-y}{y-5}\), (IV) \(-1\), and (V) \(\frac{1-y}{-10}\) are clearly not equivalent, as the numerators and/or denominators are different. As for (III) \(\frac{5-y}{2 y-10}\), simplifying the denominator results in \(\frac{5 - y}{y - 5}\), making it also equivalent to the given expression.

Key Concepts

Simplifying Algebraic Expressions

Understanding how to simplify algebraic expressions is essential in prealgebra. Simplification involves reducing an expression to its simplest form while keeping its original value unchanged. This process often includes combining like terms, using the distributive property, and canceling out terms when possible.

In the context of our example, simplifying \(\frac{3}{y-5}-\frac{y-2}{y-5}\) required combining fractions with a common denominator. After combining the numerators by subtracting, we obtained a simpler form \(\frac{5 - y}{y - 5}\). Simplification is not only about making expressions shorter; it's about making them clearer and more easily comparable to other expressions, which leads us to checking for equivalency with the given answer options.

Common Denominator

Finding a common denominator is a key step when adding or subtracting algebraic fractions. A common denominator refers to a shared base that allows fractions to be combined. It's the equivalent of finding a common language for communication — without a common denominator, fractions can't be properly communicated (combined or compared).

Our original problem, \(\frac{3}{y-5}-\frac{y-2}{y-5}\), illustrates the process with a shared denominator of \(y-5\). Working with a common denominator enabled the direct subtraction of the numerators. Without this commonality, the process of combining these fractions would be significantly more complex, involving finding the least common denominator before being able to proceed.

Algebraic Fractions

Algebraic fractions are fractions where the numerator, the denominator, or both include algebraic expressions. Handling algebraic fractions requires care as one deals not just with numbers but variables as well.

Just as with numerical fractions, the goal with algebraic fractions is often to simplify them or to combine them through operations such as addition and subtraction. Importantly, simplifying an algebraic fraction can often result in an expression that appears quite different but is in fact equivalent. This is illustrated by option (III) \(\frac{5-y}{2 y-10}\), which, upon simplifying the denominator \(2 y-10\) to \(y-5\), reveals an equivalent expression to our original algebraic fraction \(\frac{5 - y}{y - 5}\). Recognizing equivalent forms is a valuable skill when working with algebraic equations and expressions.