Problem 12 - Algebra 2 and Trigonometry

Question

Simplify each complex rational expression. In each case, list any values of the variables for which the fractions are not defined. \(\frac{y+\frac{1}{2}}{2 y+1}\)

Step-by-Step Solution

Answer

The simplified expression is \(\frac{1}{2}\), and it is undefined for \(y = -\frac{1}{2}\).

Step 1: Identify the Expression

The given expression to simplify is \(\frac{y+\frac{1}{2}}{2y+1}\). This is a complex rational expression because it involves a fraction in the numerator.

Step 2: Simplify the Numerator Expression

The numerator is \(y + \frac{1}{2}\). This can be expressed as a single fraction. Rewrite \(y\) as \(\frac{2y}{2}\) to have a common denominator: \(\frac{2y}{2} + \frac{1}{2} = \frac{2y + 1}{2}\).

Step 3: Complete the Simplification

Update the original expression, substituting the simplified numerator: \(\frac{\frac{2y + 1}{2}}{2y+1}\). Now, treat this as a division of fractions: \(\frac{2y+1}{2} \div (2y+1)\).

Step 4: Divide the Fractions

To divide the fractions, multiply by the reciprocal of the divisor: \(\frac{2y+1}{2} \times \frac{1}{2y+1}\). These expressions have a common term \(2y+1\) which cancels each other.

Step 5: Cancel the Common Factor

After canceling \(2y+1\), you are left with: \(\frac{1}{2}\). This is the simplified form of the expression.

Step 6: Identify Undefined Values

The original expression \(2y+1\) cannot be equal to zero as it appears in the denominator, as this would make the expression undefined. Solve \(2y + 1 = 0\) to find the exception: \(2y = -1\), so \(y = -\frac{1}{2}\). Therefore, \(y = -\frac{1}{2}\) is the value for which the expression is not defined.

Key Concepts

Undefined Values

Rational expressions, much like fractions, are undefined when their denominators are zero. It's very important to identify these undefined values before simplifying an expression. In our given expression \[\frac{y+\frac{1}{2}}{2y+1}\], we have a potential undefined value when the denominator is zero—that is \(2y+1=0\).
Solving this, we get:

• \(2y + 1 = 0\)
• \(2y = -1\)
• \(y = -\frac{1}{2}\)

This means the expression is undefined when \(y = -\frac{1}{2}\). Being aware of these exceptions will help avoid division by zero, which is not allowed in mathematics.

Simplifying Rational Expressions

Simplifying rational expressions involves breaking them down into their simplest form. A rational expression works similarly to a fraction where both the numerator and the denominator are polynomials.
For the expression given as \(\frac{y+\frac{1}{2}}{2y+1}\), we notice the numerator itself is a small rational expression with the addition of a fraction. You rewrite \(y\) as \(\frac{2y}{2}\) to achieve a common denominator, transforming the numerator into \(\frac{2y}{2} + \frac{1}{2} = \frac{2y + 1}{2}\).
This creates a perfectly aligned and simplified part of the entire expression before handling division.

Division of Fractions

Dividing fractions is a key skill when working with complex rational expressions. Once we expressed the numerator in the rational format \(\frac{2y + 1}{2}\), our complex rational expression became \(\frac{\frac{2y+1}{2}}{2y+1}\).
Division in fractions involves multiplying by the reciprocal of the divisor. Here, the divisor is \(2y+1\) which can be expressed as a fraction \(\frac{2y+1}{1}\). So, we write this step as \(\frac{2y+1}{2} \div \frac{2y+1}{1} = \frac{2y+1}{2} \times \frac{1}{2y+1}\).This operation helps simplify our expression significantly.

Common Denominator

Finding a common denominator is a useful technique in simplifying rational expressions. It allows fractions to be added or compared more easily. In this problem, it first appeared when simplifying \(y+\frac{1}{2}\) to \(\frac{2y+1}{2}\).
The concept of a common denominator also played a role when canceling out shared factors in the numerator and the denominator. After rewriting and multiplying by the reciprocal, \(\frac{2y+1}{2} \times \frac{1}{2y+1}\), the \(2y+1\) terms in the numerator and the denominator canceled each other out.This cancellation left us with \(\frac{1}{2}\), demonstrating a completed simplification to identify our reduced expression.